The size of arrays in Java cannot be changed. So, technically you cannot remove any elements from the array.
One way to simulate removing an element from the array is to create a new, smaller array, and then copy all of the elements from the original array into the new, smaller array.
String[] yourArray = Arrays.copyOfRange(oldArr, 1, oldArr.length);
However, I would not suggest the above method. You should really be using a List<String>. Lists allow you to add and remove items from any index. That would look similar to the following:
List<String> list = new ArrayList<String>(); // or LinkedList<String>();
list.add("Stuff");
// add lots of stuff
list.remove(0); // removes the first item
The size of arrays in Java cannot be changed. So, technically you cannot remove any elements from the array.
One way to simulate removing an element from the array is to create a new, smaller array, and then copy all of the elements from the original array into the new, smaller array.
String[] yourArray = Arrays.copyOfRange(oldArr, 1, oldArr.length);
However, I would not suggest the above method. You should really be using a List<String>. Lists allow you to add and remove items from any index. That would look similar to the following:
List<String> list = new ArrayList<String>(); // or LinkedList<String>();
list.add("Stuff");
// add lots of stuff
list.remove(0); // removes the first item
Simplest way is probably as follows - you basically need to construct a new array that is one element smaller, then copy the elements you want to keep to the right positions.
int n=oldArray.length-1;
String[] newArray=new String[n];
System.arraycopy(oldArray,1,newArray,0,n);
Note that if you find yourself doing this kind of operation frequently, it could be a sign that you should actually be using a different kind of data structure, e.g. a linked list. Constructing a new array every time is an O(n) operation, which could get expensive if your array is large. A linked list would give you O(1) removal of the first element.
An alternative idea is not to remove the first item at all, but just increment an integer that points to the first index that is in use. Users of the array will need to take this offset into account, but this can be an efficient approach. The Java String class actually uses this method internally when creating substrings.
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From my understanding the reason why removing the last element is O(1) is because you don't need to shift the array in memory. You simply remove the last element and leave the old space empty. So why is it that if you remove the first element that the Array HAS to to shift in memory (making it O(n))?
I don't understand the reasoning, if we are okay with leaving empty space in memory at the end of an array and not shifting all the other things surrounding the array in memory. Then why do we have to shift the array in memory if there is space that the start?
I am not understanding, if it's because the memory is trying to stay compact and no empty spaces are allowed. Then why don't all the other stuff in memory be shifted to the left after new space was cleared once we removed the last element from the array?
This method supports the same functionality as Collection.remove() which is how an ArrayList removes the first matching element.
public boolean remove(int n) {
for (int i = 0; i < size; i++) {
if (array[i] != n) continue;
size--;
System.arraycopy(array, i + 1, array, i, size - i);
return true;
}
return false;
}
Rather than write this code yourself, I suggest you look at Trove4J's TIntArrayList which is a wrapper for int[] You can also read the code for ArrayList to see how it is written.
You could do this:
int count; //No of elements in the array
for(i=0;i<count;i++)
{
if(Array[i]==element )
{
swap(Array,i,count);
if(count)
--count;
break;
}
}
int swap(int Array[],int i,int count)
{
int j;
for(j=i;j<=count-i;j++)
a[i]=a[i+1];
}
This is not the Full Implementation.You have to create a class and do this.
is there an array function to remove the first element of an array and return the smaller array? I am making a little shooter and I want it to be able to keep track of multiple bullets at a time.
Try this method:
Java 8+
public static <T> void remove1stArrayElementFromEveryElement(List<T[]> list) {
list.replaceAll(array -> Arrays.copyOfRange(array, 1, array.length));
}
Java 7+
public static <T> void remove1stArrayElementFromEveryElement(List<T[]> list) {
for (ListIterator<T[]> iter = list.listIterator(); iter.hasNext(); ) {
T[] array = iter.next();
array = Arrays.copyOfRange(array, 1, array.length);
iter.set(array);
}
}
I've made it generic, so that it works equally well with double[][] or int[][]. It's a void method because it directly modifies the passed list, without making a copy. The only copy it makes is a temporary one of each array that's in the list. Also, it works with any list, not just ArrayList.
The size of arrays in Java cannot be changed. So, you cannot technically remove an element from an array. You basically need to construct a new array and then copy from the original. This is exactly what you did.
For copying, Arrays.copyOfRange(oldArr, 1, oldArr.length) maybe the easiest way. Try it to see there is any change.
I think you have some other options here:
Increase Heap (with -Xmx1024m for example)
Reconsider you double[][] input? Is there any work around?
If the comparison for "remove first" is between the ArrayList and the LinkedList classes, the LinkedList wins clearly.
Removing an element from a linked list costs O(1), while doing so for an array (array list) costs O(n).
I think that what you need is an ArrayDeque (an unfairly overlooked class in java.util). Its removeFirst method performs in O(1) as for LinkedList, while it generally shows the better space and time characteristics of ArrayList. It’s implemented as a circular queue in an array.
You should very rarely use LinkedList. I did once in my 17 years as Java programmer and regretted in retrospect.