How to loop through ArrayList for different values in java

Question

I am just practising java.

I have this ArrayList which contains duplicate values.

I want to loop through this list so that it can print these duplicate values by group.

here is what my method contains.

ArrayList nomero = new ArrayList();

    nomero.add(1);
    nomero.add(1);
    nomero.add(2);
    nomero.add(2);
    nomero.add(3);
    nomero.add(3);


    for(int i=0; i<nomero.size(); i++){

        System.out.println(nomero.get(i));

    }

and the it prints like below.

what I wanna do is print it like

Please help me with this issue. Thanks.

if value != lastValue print new line so basically you need to know the last "group" value and when it changes, you print a new line — MadProgrammer
– MadProgrammer, Commented Jan 28, 2017 at 1:48

Elliott Frisch · Accepted Answer · 2017-01-28 01:50:57Z

3

Simple enough, add a line break when the previous value is different from the current. But, you shouldn't use raw types. And you should sort the List (and prefer the interface to the concrete type). And you might use Arrays.asList to initialize your List. Something like,

List<Integer> nomero = new ArrayList<>(Arrays.asList(1, 1, 2, 2, 3, 3));
Collections.sort(nomero);
for (int i = 0; i < nomero.size(); i++) {
    if (i > 0 && nomero.get(i - 1) != nomero.get(i)) {
        System.out.println();
    }
    System.out.println(nomero.get(i));
}

answered Jan 28, 2017 at 1:50

Elliott Frisch

202k20 gold badges166 silver badges265 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

user1971804 Over a year ago

Great answers, I was going mad to figure out this. Simple and explanatory answers. Thanks a lot everyone.

MadProgrammer · Accepted Answer · 2017-01-28 01:51:47Z

1

So, at a very basic level, you need to know what the "last value" was, when it longer equals the "current value", you add a new line and update the "last value" to be equal to the "current value"

ArrayList<Integer> nomero = new ArrayList<>();

nomero.add(1);
nomero.add(1);
nomero.add(2);
nomero.add(2);
nomero.add(3);
nomero.add(3);

int lastValue = nomero.get(0);

for (int i = 0; i < nomero.size(); i++) {
    int nextValue = nomero.get(i);
    if (nextValue != lastValue) {
        System.out.println("");
        lastValue = nextValue;
    }
    System.out.println(nextValue);

}

This does assume that the list is already sorted

answered Jan 28, 2017 at 1:51

community wiki

MadProgrammer

Comments

user3325637 · Accepted Answer · 2017-01-31 10:40:09Z

0

Java 8 made it simpler ...

List<Integer> nomero = Arrays.asList(1, 1, 2, 2, 3, 3);
Map<Integer, Long> map = nomero.stream().collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));
    for (Integer i : map.keySet()) {
        for (int j = 0; j < map.get(i); j++) {
            System.out.println(i);
        }
        System.out.println();
    }

answered Jan 31, 2017 at 10:40

user3325637

1241 silver badge4 bronze badges

Collectives™ on Stack Overflow

How to loop through ArrayList for different values in java

3 Answers 3

1 Comment

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related