In your case, there's no need to iterate through the list, because you know which object to delete. You have several options. First you can remove the object by index (so if you know, that the object is the second list element):
a.remove(1); // indexes are zero-based
Or, you can remove the first occurence of your string:
a.remove("acbd"); // removes the first String object that is equal to the
// String represented by this literal
Or, remove all strings with a certain value:
while(a.remove("acbd")) {}
It's a bit more complicated, if you have more complex objects in your collection and want to remove instances, that have a certain property. So that you can't remove them by using remove with an object that is equal to the one you want to delete.
In those case, I usually use a second list to collect all instances that I want to delete and remove them in a second pass:
List<MyBean> deleteCandidates = new ArrayList<>();
List<MyBean> myBeans = getThemFromSomewhere();
// Pass 1 - collect delete candidates
for (MyBean myBean : myBeans) {
if (shallBeDeleted(myBean)) {
deleteCandidates.add(myBean);
}
}
// Pass 2 - delete
for (MyBean deleteCandidate : deleteCandidates) {
myBeans.remove(deleteCandidate);
}
Answer from Andreas Dolk on Stack Overflowjava - How to remove element from ArrayList by checking its value? - Stack Overflow
Question and solution: How to delete or add an object to/from array list while iterating it.
How does an ArrayList remove an element from an array?
Why is the add(index, element) time complexity not constant in Java for array lists?
For array lists, you would have to move all elements to adjacent positions when you insert at an index. So it takes linear time ( more the number of elements already in the list, more time it takes to move them all ).
Also Java has nothing to do with time complexities of a data structure. It is universal.
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In your case, there's no need to iterate through the list, because you know which object to delete. You have several options. First you can remove the object by index (so if you know, that the object is the second list element):
a.remove(1); // indexes are zero-based
Or, you can remove the first occurence of your string:
a.remove("acbd"); // removes the first String object that is equal to the
// String represented by this literal
Or, remove all strings with a certain value:
while(a.remove("acbd")) {}
It's a bit more complicated, if you have more complex objects in your collection and want to remove instances, that have a certain property. So that you can't remove them by using remove with an object that is equal to the one you want to delete.
In those case, I usually use a second list to collect all instances that I want to delete and remove them in a second pass:
List<MyBean> deleteCandidates = new ArrayList<>();
List<MyBean> myBeans = getThemFromSomewhere();
// Pass 1 - collect delete candidates
for (MyBean myBean : myBeans) {
if (shallBeDeleted(myBean)) {
deleteCandidates.add(myBean);
}
}
// Pass 2 - delete
for (MyBean deleteCandidate : deleteCandidates) {
myBeans.remove(deleteCandidate);
}
One-liner (java8):
list.removeIf(s -> s.equals("acbd")); // removes all instances, not just the 1st one
(does all the iterating implicitly)