Question

the code below compiles perfectly for 5 integers entered by the user.

what i want to do is alter my code so that i can as the user how many numbers the user wants in the list. then the program will ask the user to enter that many integers and then the program will find the largest of those integers. can anyone help me alter the code below to fit these new standards?

{
    int integer1 = 
    Integer.parseInt(JOptionPane.showInputDialog( "Enter an integer:" ));
    int integer2 = 
    Integer.parseInt(JOptionPane.showInputDialog( "Enter an integer:" ));
    int integer3 = 
    Integer.parseInt(JOptionPane.showInputDialog( "Enter an integer:" ));
    int integer4 = 
    Integer.parseInt(JOptionPane.showInputDialog( "Enter an integer:" ));
    int integer5 = 
    Integer.parseInt(JOptionPane.showInputDialog( "Enter an integer:" ));   

    if (integer1 > integer2 && integer1 > integer3 && integer1 > integer4 && integer1 > integer5)
        JOptionPane.showMessageDialog(null, "The largest number is: " + integer1);
    if (integer2 > integer1 && integer2 > integer3 && integer2 > integer4 && integer2 > integer5)
        JOptionPane.showMessageDialog(null, "The largest number is: " + integer2);
    if (integer3 > integer2 && integer3 > integer1 && integer3 > integer4 && integer3 > integer5)
        JOptionPane.showMessageDialog(null, "The largest number is: " + integer3);
    if (integer4 > integer1 && integer4 > integer3 && integer4 > integer2 && integer4 > integer5)
        JOptionPane.showMessageDialog(null, "The largest number is: " + integer4);
    if (integer5 > integer1 && integer5 > integer3 && integer5 > integer4 && integer5 > integer2)
        JOptionPane.showMessageDialog(null, "The largest number is: " + integer5);
 }

i first want to prompt the user with

how many integers do you want in your list?

then i want to say

enter an integer: 

as many times as the user said he wanted.

Answer 1

The key here is to use Collections.max.

Returns the maximum element of the given collection, according to the natural ordering of its elements.

The natural ordering for Integer is from least-to-greatest, i.e. ascending. This makes it perfect to use here.

int largest = Collections.max(Arrays.asList(integer1, integer2, integer3,
    integer4, integer5));

Alternatively, you could just build the List using a loop instead. See below for code that prompts the user to input the number of integers to enter.

int n = Integer.parseInt(
    JOptionPane.showInputDialog("How many integers do you want in your list?"));
List<Integer> inputs = new ArrayList<Integer>(n);
for (int i = 0; i < n; ++i) {
  inputs.add(Integer.parseInt(
      JOptionPane.showInputDialog("Enter an integer:")));
}
int largest = Collections.max(inputs);
JOptionPane.showMessageDialog(null, "The largest number is: " + largest);