List has the method add(int, E), so you can use:
list.add(0, yourObject);
Afterwards you can delete the last element with:
if(list.size() > 10)
list.remove(list.size() - 1);
However, you might want to rethink your requirements or use a different data structure, like a Queue
EDIT
Maybe have a look at Apache's CircularFifoQueue:
CircularFifoQueueis a first-in first-out queue with a fixed size that replaces its oldest element if full.
Just initialize it with you maximum size:
CircularFifoQueue queue = new CircularFifoQueue(10);
arrays - Java ArrayList how to add elements at the beginning - Stack Overflow
Is there a way to add elements to an Array List all at once?
java - Adding values to Arraylist - Stack Overflow
java - Add object to ArrayList at specified index - Stack Overflow
List has the method add(int, E), so you can use:
list.add(0, yourObject);
Afterwards you can delete the last element with:
if(list.size() > 10)
list.remove(list.size() - 1);
However, you might want to rethink your requirements or use a different data structure, like a Queue
EDIT
Maybe have a look at Apache's CircularFifoQueue:
CircularFifoQueueis a first-in first-out queue with a fixed size that replaces its oldest element if full.
Just initialize it with you maximum size:
CircularFifoQueue queue = new CircularFifoQueue(10);
Using Specific Datastructures
There are various data structures which are optimized for adding elements at the first index. Mind though, that if you convert your collection to one of these, the conversation will probably need a time and space complexity of O(n)
Deque
The JDK includes the Deque structure which offers methods like addFirst(e) and offerFirst(e)
Deque<String> deque = new LinkedList<>();
deque.add("two");
deque.add("one");
deque.addFirst("three");
//prints "three", "two", "one"
Analysis
Space and time complexity of insertion is with LinkedList constant (O(1)). See the Big-O cheatsheet.
Reversing the List
A very easy but inefficient method is to use reverse:
Collections.reverse(list);
list.add(elementForTop);
Collections.reverse(list);
If you use Java 8 streams, this answer might interest you.
Analysis
- Time Complexity:
O(n) - Space Complexity:
O(1)
Looking at the JDK implementation this has a O(n) time complexity so only suitable for very small lists.
For example I can do this with a basic integer array:
int[] test1 = new int[]{2, 1, 5, 1, 3, 2};But with an Array list I have to add them one at a time like this
List<Integer> test1 = new ArrayList<Integer>(); test1.add(2); test1.add(1); test1.add(3);
?
First simple rule: never use the String(String) constructor, it is absolutely useless (*).
So arr.add("ss") is just fine.
With 3 it's slightly different: 3 is an int literal, which is not an object. Only objects can be put into a List. So the int will need to be converted into an Integer object. In most cases that will be done automagically for you (that process is called autoboxing). It effectively does the same thing as Integer.valueOf(3) which can (and will) avoid creating a new Integer instance in some cases.
So actually writing arr.add(3) is usually a better idea than using arr.add(new Integer(3)), because it can avoid creating a new Integer object and instead reuse and existing one.
Disclaimer: I am focusing on the difference between the second and third code blocks here and pretty much ignoring the generics part. For more information on the generics, please check out the other answers.
(*) there are some obscure corner cases where it is useful, but once you approach those you'll know never to take absolute statements as absolutes ;-)
The second one would be preferred:
- it avoids unnecessary/inefficient constructor calls
- it makes you specify the element type for the list (if that is missing, you get a warning)
However, having two different types of object in the same list has a bit of a bad design smell. We need more context to speak on that.
You can do it like this:
list.add(1, object1)
list.add(2, object3)
list.add(2, object2)
After you add object2 to position 2, it will move object3 to position 3.
If you want object3 to be at position3 all the time I'd suggest you use a HashMap with position as key and object as a value.
You can use Array of objects and convert it to ArrayList-
Object[] array= new Object[10];
array[0]="1";
array[3]= "3";
array[2]="2";
array[7]="7";
List<Object> list= Arrays.asList(array);
ArrayList will be- [1, null, 2, 3, null, null, null, 7, null, null]