🔒 Closed HELP MGA MASTER! JAVA

Status
Not open for further replies.
loop mo lang yan

loop You do not have permission to view the full content of this post. Log in or register now.

if even / odd You do not have permission to view the full content of this post. Log in or register now.

sa 2nd

loop lang din

tas gawa ka variable lowest and highest

tas sa highest

if number now is > number stored
--- set mo yung var to number now

same sa lowenst
 
Pa-test nalang din po, then paki check na din ng array.
[CODE lang="java" title="Problem 1"]public class Main {
public static void main(String[] args) {
int[] field = {2, 20, 3, 13, 15, 3, 9, 10, 1, 5, 8};
int oddCount = 0;
for(int e : field)
if(e % 2 != 0)
oddCount++;
System.out.println("Odd Numbers Count: " + oddCount);
}
}[/CODE]
 
[CODE lang="java" title="Problem 2"]public class Main {
public static void main(String[] args) {
int[] field = {2, 20, 3, 13, 15, 3, 9, 10, 1, 5, 8};
bubbleSort(field);
System.out.println("Max Number in Field " + field[field.length-1]);
System.out.println("Min Number in Field " + field[0]);
}

static int[] bubbleSort(int[] array){
for(int i = 0; i < array.length; i++){
for(int j = 0; j < array.length - 1; j++){
if(array[j] > array[j + 1]){
int temp = array[j];
array[j] = array[j + 1];
array[j + 1] = temp;
}
}
}
return array;
}
}[/CODE]
 
Status
Not open for further replies.

About this Thread

  • 7
    Replies
  • 377
    Views
  • 4
    Participants
Last reply from:
Arcturus

Trending Topics

Online now

Members online
1,112
Guests online
1,690
Total visitors
2,802

Forum statistics

Threads
2,295,002
Posts
29,094,727
Members
1,206,648
Latest member
flian5y89
Back
Top