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, Aug 12 '15 at 09:20
If there is more than 12 elements, should the first 12 still be printed? — Ian2thedv, Aug 12 '15 at 09:39
possible duplicate of [Ways to iterate over a List in java?](http://stackoverflow.com/questions/18410035/ways-to-iterate-over-a-list-in-java) — Piyush Mittal, Aug 12 '15 at 09:48

score 2 · Answer 1 · edited Aug 12 '15 at 09:35

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 '15 at 09:35

vefthym

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answered Aug 12 '15 at 09:21

Suresh Atta

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score 2 · Answer 2 · answered Aug 12 '15 at 09:22

2

This is a good use-case for streams.

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

answered Aug 12 '15 at 09:22

Peter Lawrey

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score 1 · Answer 3 · answered Aug 12 '15 at 09:26

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 '15 at 09:26

Erwan C.

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Well, `Math.min` looks much better then this ternary operator :P. – Tom Aug 12 '15 at 10:11

score 0 · Answer 4 · answered Aug 12 '15 at 09:31

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 

 }     
    }

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 — Mark Fisher, Aug 12 '15 at 10:14

score 0 · Answer 5 · answered Aug 12 '15 at 09:32

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 '15 at 09:32

Uma Kanth

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4

This is not optimal as this compute the min on each loop. – Vouze Aug 12 '15 at 09:37
list.size() is O(1) (see http://stackoverflow.com/questions/6540511/time-complexity-for-java-arraylist) and Math.min(num, num) is hardly going to add any significant overhead. – Mark Fisher Aug 12 '15 at 10:12
Also Math.min() is same as the `<` operator, has no complexity. Have a look at [this](http://stackoverflow.com/questions/2103606/which-one-is-faster-in-java-1-math-maxa-b-2-abab). ThIs keeps the for statement clean. – Uma Kanth Aug 12 '15 at 10:20
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 Aug 12 '15 at 14:20

Vouze · Answer 6 · 2015-08-12T14:39:33.867

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;
            }
        }
    }

Java List looping based of available size

6 Answers6