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Title
# announcements
m
Marcin Wisniowski
11/10/2019, 11:49 AMDoes anyone have an idiomatic way of checking if the last
nlist.takeLast(n).toSet().size == 1k
Kroppeb
11/10/2019, 12:08 PMtoSet()O(n)m
Marcin Wisniowski
11/10/2019, 12:14 PMTrue, maybe
list.takeLast(n).all { it == list.last() }But it still unnecessarily compares the last element to itself.
k
Kroppeb
11/10/2019, 12:15 PMYou could put it in an extension function if it's a thing you need more often.
If you really hate the fact it compares the last element with itself, you could take a sublist that doesn't contain the last element, but it won't make that much of a difference.
m
molikuner
11/10/2019, 1:51 PMlist.takeLast(n).dropLast(1).all { it == list.last() }t
todd.ginsberg
11/10/2019, 2:26 PMlist.takeLast(n).distinct().size == 1k
karelpeeters
11/10/2019, 2:38 PMExcept that's slower than
toSet()distinct() = toSet().toList()t
todd.ginsberg
11/10/2019, 2:39 PMYeah, that's fair.
It's easier to read though, I feel.
👍 1
p
pablisco
11/10/2019, 8:30 PMYou could overload the default
all Copy code
fun <T> List<T>.all(from: Int = 0, f: (T, T) -> Boolean): Boolean {
for(i in from until (size-1)) if(!f(get(i), get(i + 1))) return false;
return true
} Copy code
list.all(from = n) { a, b -> a == b }d
Derek Peirce
11/11/2019, 1:26 AMTry
!list.takeLast(n).asSequence().distinct().drop(1).any()