See the javadoc
of List
list.get(0);
or Set
set.iterator().next();
and check the size before using the above methods by invoking isEmpty()
!list_or_set.isEmpty()
Answer from stacker on Stack OverflowSee the javadoc
of List
list.get(0);
or Set
set.iterator().next();
and check the size before using the above methods by invoking isEmpty()
!list_or_set.isEmpty()
Collection c;
Iterator iter = c.iterator();
Object first = iter.next();
(This is the closest you'll get to having the "first" element of a Set. You should realize that it has absolutely no meaning for most implementations of Set. This may have meaning for LinkedHashSet and TreeSet, but not for HashSet.)
playersList.get(0)
Java has limited operator polymorphism. So you use the get() method on List objects, not the array index operator ([])
You have to access lists a little differently than arrays in Java. See the javadocs for the List interface for more information.
playersList.get(0)
However if you want to find the smallest element in playersList, you shouldn't sort it and then get the first element. This runs very slowly compared to just searching once through the list to find the smallest element.
For example:
int smallestIndex = 0;
for (int i = 1; i < playersList.size(); i++) {
if (playersList.get(i) < playersList.get(smallestIndex))
smallestIndex = i;
}
playersList.get(smallestIndex);
The above code will find the smallest element in O(n) instead of O(n log n) time.