Linked lists are preferable over arrays when: you need constant-time insertions/deletions from the list (such as in real-time computing where time predictability is absolutely critical) you don’t know how many items will be in the list. With arrays, you may need to re-declare and copy memory if the array grows too big you don’t need … Read more
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: CircularFifoQueue is a first-in … Read more
You can use subList(int fromIndex, int toIndex) to get a view of a portion of the original list. From the API: Returns a view of the portion of this list between the specified fromIndex, inclusive, and toIndex, exclusive. (If fromIndex and toIndex are equal, the returned list is empty.) The returned list is backed by … Read more
a < b and a – b < 0 can mean two different things. Consider the following code: int a = Integer.MAX_VALUE; int b = Integer.MIN_VALUE; if (a < b) { System.out.println(“a < b”); } if (a – b < 0) { System.out.println(“a – b < 0”); } When run, this will only print a … Read more
Check out the set(int index, E element) method in the List interface
Use Collections.shuffle() twice, with two Random objects initialized with the same seed: long seed = System.nanoTime(); Collections.shuffle(fileList, new Random(seed)); Collections.shuffle(imgList, new Random(seed)); Using two Random objects with the same seed ensures that both lists will be shuffled in exactly the same way. This allows for two separate collections.
Yes, assignment will just copy the value of l1 (which is a reference) to l2. They will both refer to the same object. Creating a shallow copy is pretty easy though: List<Integer> newList = new ArrayList<>(oldList); (Just as one example.)
Two options: Create a list of values you wish to remove, adding to that list within the loop, then call originalList.removeAll(valuesToRemove) at the end Use the remove() method on the iterator itself. Note that this means you can’t use the enhanced for loop. As an example of the second option, removing any strings with a … Read more
Not every List implementation supports the add() method. One common example is the List returned by Arrays.asList(): it is documented not to support any structural modification (i.e. removing or adding elements) (emphasis mine): Returns a fixed-size list backed by the specified array. Even if that’s not the specific List you’re trying to modify, the answer … Read more
Arrays.asList can help here: new ArrayList<Integer>(Arrays.asList(1,2,3,5,8,13,21));