Java List looping based of available size

Question

I want to display only 12 sorted elements from myList. If the size is less than 12, say 5, or 3, or 1, still I need to loop and display only those available items.

Below is my code:

public class JavaApplication {

    public static void main(String[] args) {

        List<String> myList = new ArrayList<>();
        myList.add("20150830");
        myList.add("20141201");
        myList.add("20150716");
        myList.add("20151131"); 
        myList.add("20141101");
        myList.add("20150620");
        myList.add("20150301");
        myList.add("20150702");
        myList.add("20150511");

        Collections.sort(myList,Collections.reverseOrder());

        for(int i = 0; i < myList.size(); i++) {
            System.out.println(myList.get(i).toString());
        }
    }     
}

Alternatively for(int i = 0; i < Math.min(myList.size(),12); i++){ — Florian Schaetz
– Florian Schaetz, Commented Aug 12, 2015 at 9:20
If there is more than 12 elements, should the first 12 still be printed? — Ian2thedv
– Ian2thedv, Commented Aug 12, 2015 at 9:39

Peter Lawrey · Accepted Answer · 2015-08-12 09:22:55Z

2

This is a good use-case for streams.

myList.stream()
        .sorted(Comparator.<String>reverseOrder())
        .limit(12)
        .forEach(System.out::println);

answered Aug 12, 2015 at 9:22

Peter Lawrey

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Comments

vefthym · Accepted Answer · 2015-08-12 09:35:08Z

2

Just add one more condition in your loop to restrict the loop for 12.

for(int i = 0; i < myList.size() &&  i < 12 ; i++){
    System.out.println(myList.get(i).toString());
}

edited Aug 12, 2015 at 9:35

vefthym

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answered Aug 12, 2015 at 9:21

Suresh Atta

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Comments

Erwan C. · Accepted Answer · 2015-08-12 09:26:09Z

1

You can try:

int maxValuesToDisplay=12;
int maxToIterate=(myList.size()<maxValuesToDisplay) ? myList.size() : maxValuesToDisplay;

for(int i = 0; i < maxToIterate; i++){
    System.out.println(myList.get(i).toString());
}

answered Aug 12, 2015 at 9:26

Erwan C.

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1 Comment

Tom Over a year ago

Well, Math.min looks much better then this ternary operator :P.

Kumar Abhinav · Accepted Answer · 2015-08-12 09:31:02Z

0

Use List.subList(int fromIndex, int toIndex);

public class JavaApplication {

    public static void main(String[] args) {

        List<String> myList = new ArrayList<>();
        myList.add("20150830");
        myList.add("20141201");
        myList.add("20150716");
        myList.add("20151131"); 
        myList.add("20141101");
        myList.add("20150620");
        myList.add("20150301");
        myList.add("20150702");
        myList.add("20150511");


        Collections.sort(myList,Collections.reverseOrder());
        int a = myList.size();
        List subList = null;
        if(a<12)
        subList = myList.subList(0,a);
        else subList = myList.subList(0,12);
       //Print sublist now 

 }     
    }

answered Aug 12, 2015 at 9:31

Kumar Abhinav

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2 Comments

Mark Fisher Over a year ago

Combine this with Math.min() would be much cleaner: List subList = myList.subList(0, Math.min(myList.size(), 12)), all in 1 line, not 5

Kumar Abhinav Over a year ago

@MarkFisher True Mark :)

Uma Kanth · Accepted Answer · 2015-08-12 09:32:56Z

0

You could use the Math.min() function and iterate the minimum of 12 and the list-size.

for(int i = 0; i < Math.min(myList.size(), 12); i++){
    System.out.println(myList.get(i).toString());
   }

answered Aug 12, 2015 at 9:32

Uma Kanth

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4 Comments

Vouze Over a year ago

This is not optimal as this compute the min on each loop.

Mark Fisher Over a year ago

list.size() is O(1) (see stackoverflow.com/questions/6540511/…) and Math.min(num, num) is hardly going to add any significant overhead.

Uma Kanth Over a year ago

Also Math.min() is same as the < operator, has no complexity. Have a look at this. ThIs keeps the for statement clean.

Vouze Over a year ago

Except this is an object oriented language and you don't know what is behind "size()". Anyway for optimization, always get out of the loops invariants you can compute.

Vouze · Accepted Answer · 2015-08-12 14:39:33Z

You can use TreeSet :

public static void main(String[] args) {
    Set<String> myList = new TreeSet<>();
    myList.add("20150830");
    myList.add("20141201");
    myList.add("20150716");
    myList.add("20151131");
    myList.add("20141101");
    myList.add("20150620");
    myList.add("20150301");
    myList.add("20150702");
    myList.add("20150511");

    int i = 0;
    for(String s : myList){
        System.out.println(s);
        i++;
        if(i >= 5) {
            break;
        }
    }
}

And for reverse order :

    public static void main(String[] args) {
        TreeSet<String> myList = new TreeSet<>();
        myList.add("20150830");
        myList.add("20141201");
        myList.add("20150716");
        myList.add("20151131");
        myList.add("20141101");
        myList.add("20150620");
        myList.add("20150301");
        myList.add("20150702");
        myList.add("20150511");

        Iterator<String> it = myList.descendingIterator();
        int i = 0;

        while(it.hasNext()) {
            String s = it.next();
            System.out.println(s);
            i++;
            if (i >= 5) {
                break;
            }
        }
    }

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