CodeGym/Help with Java Tasks/can someone help?

Level 4

18.11.2018
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15comments

can someone help?

Question about the task

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the program is all of it right but i had a problem with the last condition, i tried in many ways but i did not found the right one.... What should the code be to do the last condition?

package com.codegym.task.task04.task0419; /* Maximum of four numbers */ import java.io.*; public class Solution { public static void main(String[] args) throws Exception { BufferedReader reader = new BufferedReader(new InputStreamReader (System.in)); String numo = reader.readLine(); int num1 = Integer.parseInt(numo); String numT = reader.readLine(); int num2 = Integer.parseInt(numT); String numt = reader.readLine(); int num3 = Integer.parseInt(numt); String numf = reader.readLine(); int num4 = Integer.parseInt(numf); if(num1 > num2 && num1 > num3 && num1 > num4){System.out.println(num1);} if(num2 > num1 && num2 > num3 && num2 > num4){System.out.println(num2);} if(num3 > num1 && num3 > num2 && num3 > num4){System.out.println(num3);} if(num4 > num1 && num4 > num2 && num4 > num3){System.out.println(num4);} if(num1 == num2 && num2 == num3 && num3 == num4 || num2 == num1 && num2 == num3 && num2 == num4 ||num3 == num1 && num3 == num2 && num3 == num4 ||num4 == num1 && num4 == num3 && num4 == num2 ){System.out.println(num1);} } }

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Suma

Level 11 , Hyderabad, India

31 May 2019, 08:45

Using Math.max function to get the maximum number /* Maximum of four numbers */ import java.io.*; public class Solution { public static void main(String[] args) throws Exception { //write your code here BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); int a = Integer.parseInt(reader.readLine()); int b = Integer.parseInt(reader.readLine()); int c = Integer.parseInt(reader.readLine()); int d = Integer.parseInt(reader.readLine()); System.out.print(Math.max(Math.max(a,b),Math.max(c,d))); } }

hidden #10444738

Level 16

4 May 2019, 14:50

I used Nikets code and inserted this on the first if statement. if (x>=y || x == y) {

Thomas

Level 12 , Scottsdale, United States

26 May 2019, 19:19

Yeah .... not sure why yet ... thinking

hidden #10444738

Level 16

4 May 2019, 05:45

i'm still confused as to when to use BufferedReader and when to use Scanner!

nikk_shukla

Level 10 , Mumbai, India

14 May 2019, 09:46

Read This Article:- https://www.geeksforgeeks.org/difference-between-scanner-and-bufferreader-class-in-java/

William Leininger

Level 5 , Vancouver, WA, United States

11 February 2019, 08:30

at the most basic level what the last condition is wanting/telling you to do is have a catch all final bit of code which is an "else" statement that has no requirements. if (condition) code else if (second condition) code else (no condition as it is going to execute code if the conditions before it don't execute) code review this. https://codegym.cc/quests/lectures/questsyntax.level04.lecture05

Niket

Level 8 , Kolkata, India

10 January 2019, 09:32

package com.codegym.task.task04.task0419;

/*
Maximum of four numbers

*/

import java.io.*;

public class Solution {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int a = Integer.parseInt(br.readLine());
        int b = Integer.parseInt(br.readLine());
        int c = Integer.parseInt(br.readLine());
        int d = Integer.parseInt(br.readLine());
        int m1 = maxOfTwo(a,b);
        int m2 = maxOfTwo(c,d);
        int m3 = maxOfTwo(m1,m2);
        System.out.println(m3);
    }
    public static int maxOfTwo(int x, int y) {
        if (x>=y) {
            return x;
        }
        else {
            return y;
        }
    }
}

+14

Muhammad Yasir

Level 5 , Islamabad, Pakistan

10 February 2019, 12:10

Thank you for sharing the code! Your approach is very simple, easier and compact.

Rembrandt Escultura

Level 4 , San Jose, United States

14 March 2019, 22:00

I forgot about this one. Wow. So much more efficient than my code. Thanks for posting this.

John Kluse

Level 6 , Bountiful, United States

17 March 2019, 18:13

Cool, thanks for sharing this. It looks like it is very similar to the Math.max method that takes two integers and returns the larger of the two.

Connor

Level 8 , Boston, United States

19 April 2019, 00:18

That's very cool! Completely made me rethink my approach. Does it have a way to print the numbers if they equal each other?

Thomas

Level 12 , Scottsdale, United States

26 May 2019, 19:17

Thanks Niket --- really clean ! like the use of Max method / function -- trying to use more functions to apply the whole language & reduce statement programming ...

romeo

Level 5 , Paramaribo, Suriname

28 May 2019, 20:43

Gadamn! that's beautiful

Mohamad

Level 4

18 November 2018, 18:25

thanx bro

Michael Martin

Level 19 , Arlington, United States

18 November 2018, 17:29

useful

try: if((num1 == num2) && (num1 == num3) && (num1 == num4)) System.out.println(num1); Because either all the numbers are different or they are all equal to each other.