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JAVA Concept Question: Finding the Larger Value

Posted on 2013-01-02
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Last Modified: 2013-01-03
Hello, I am fairly new to the Java Programming Language and have a question regarding a practice problem that I have completed. The problem that I finished is as follows.

Given 2 positive int values, return the larger value that is in the range 10..20 inclusive, or return 0 if neither is in that range.

max1020(11, 19) ¿ 19
max1020(19, 11) ¿ 19
max1020(11, 9) ¿ 11

Below is the code that I have written followed by one of the solutions that the website provided.

My Code:

public int max1020(int a, int b) {
  if ((a >= 10) && (a <= 20)){
    if ((b >= 10) && (b <= 20)){
      if (a > b){
        return a;
      }
      else return b;
    }
    else return a;
  }
  else {
    if ((b >= 10) && (b <= 20)){
      return b;
    }
    return 0;
  }
}


Solution Code:

public int max1020(int a, int b) {
  // First make it so the bigger value is in a
  if (b > a) {
    int temp = a;
    a = b;
    b = temp;
  }
 
  // Knowing a is bigger, just check a first
  if (a >= 10 && a <= 20) return a;
  if (b >= 10 && b <= 20) return b;
  return 0;
}

The question that I would like to ask is why does the solution code use the first if statement? I understand that the comment says the statement is checking for the larger value, but do not understand how it does that.

Thank you for your help.
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Comment
Question by:Paul_ATL
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3 Comments
 
LVL 16

Assisted Solution

by:krakatoa
krakatoa earned 390 total points
ID: 38738068
This might be neater

import java.util.Scanner;
import java.lang.Math.*;

class ReturnLargest
{
      public static void main (String [] args)
      {
            Scanner keyIn = new Scanner(System.in);
            
            int numA = 0;
            int numB = 0; 
	
            int largest =0;
            int smallest = 0;                      
            
                System.out.print("Enter an integer: \n");
                numA = keyIn.nextInt( );
	System.out.print("Enter an integer: \n");
                numB = keyIn.nextInt( );
                largest = Math.max(numA,numB);
	smallest = Math.min(numA,numB);
               if((largest>=10&&largest<=20)||(smallest>=10&&smallest<=20)){System.out.println("\n"+largest);}else{System.out.println("\n0");}
            	
      }
}

Open in new window


re: Your comment : if you mean this :  if (b > a) {
Both ints need to be checked for range, and you then need to know which is the largest.
0
 
LVL 16

Accepted Solution

by:
imladris earned 510 total points
ID: 38740692
In this piece:

// First make it so the bigger value is in a
  if (b > a) {
    int temp = a;
    a = b;
    b = temp;
  }

It checks a and b. If b is bigger than a it swaps the values in a and b, so that a is now bigger. So at the end of this clause, a is always bigger than b.

Then, in the next part, it only has to check for range.
0
 
LVL 16

Expert Comment

by:krakatoa
ID: 38741299
Just wondering where I or imladris or both went off the rails to get only a 'good'. The 'excellent' category doesn't cost you any extra points. But more importantly, it would be interesting to hear what you were not 100% happy with.
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