Reversing a Queue and converting it into an int array
I have a Queue<Integer>
declared as Queue<Integer> queue=new LinkedList();
, I need to reverse the elments order in it, and then convert it into an int array. I wrote below code:
Collections.reverse((List)queue);
int res=queue.stream().mapToInt(Integer::intValue).toArray();
This code has two problems:
- the explict casting
(List)queue
; - I wonder if there is a one line solution.
So do we have any more elegant way to do this?
Clearification of the problem:
Whether the queue is reversed is not important. An int array of the reversed elements is what I need.
java collections queue
add a comment |
I have a Queue<Integer>
declared as Queue<Integer> queue=new LinkedList();
, I need to reverse the elments order in it, and then convert it into an int array. I wrote below code:
Collections.reverse((List)queue);
int res=queue.stream().mapToInt(Integer::intValue).toArray();
This code has two problems:
- the explict casting
(List)queue
; - I wonder if there is a one line solution.
So do we have any more elegant way to do this?
Clearification of the problem:
Whether the queue is reversed is not important. An int array of the reversed elements is what I need.
java collections queue
add a comment |
I have a Queue<Integer>
declared as Queue<Integer> queue=new LinkedList();
, I need to reverse the elments order in it, and then convert it into an int array. I wrote below code:
Collections.reverse((List)queue);
int res=queue.stream().mapToInt(Integer::intValue).toArray();
This code has two problems:
- the explict casting
(List)queue
; - I wonder if there is a one line solution.
So do we have any more elegant way to do this?
Clearification of the problem:
Whether the queue is reversed is not important. An int array of the reversed elements is what I need.
java collections queue
I have a Queue<Integer>
declared as Queue<Integer> queue=new LinkedList();
, I need to reverse the elments order in it, and then convert it into an int array. I wrote below code:
Collections.reverse((List)queue);
int res=queue.stream().mapToInt(Integer::intValue).toArray();
This code has two problems:
- the explict casting
(List)queue
; - I wonder if there is a one line solution.
So do we have any more elegant way to do this?
Clearification of the problem:
Whether the queue is reversed is not important. An int array of the reversed elements is what I need.
java collections queue
java collections queue
edited 19 hours ago
Moira
5,25221937
5,25221937
asked yesterday
ZhaoGang
1,6181015
1,6181015
add a comment |
add a comment |
8 Answers
8
active
oldest
votes
First, please don't use raw types (do use the diamond operator). Not quite a one liner, but you could first convert to an int
and then use commons lang ArrayUtils.reverse(int)
like
Queue<Integer> queue = new LinkedList<>();
// ...
int arr = queue.stream().mapToInt(Integer::intValue).toArray();
ArrayUtils.reverse(arr);
You could also write your own int
reverse method that allowed for a fluent interface (e.g. return the int
) then you could make it a one liner. Like,
public static int reverse(int arr) {
for (int i = 0; i < arr.length / 2; i++) {
int temp = arr[i];
arr[i] = arr[arr.length - i - 1];
arr[arr.length - i - 1] = temp;
}
return arr;
}
And then
int arr = reverse(queue.stream().mapToInt(Integer::intValue).toArray());
but this wouldn't reverse the queue.
– nullpointer
yesterday
2
@nullpointer True. But, if the goal is a reversedint
then it isn't clear that the queue must also be reversed. In fact, I would assume the queue goes out of scope and theint
is returned to the caller.
– Elliott Frisch
yesterday
add a comment |
No need to get fancy here.
static int toReversedArray(Queue<Integer> queue) {
int i = queue.size();
int array = new int[i];
for (int element : queue) {
array[--i] = element;
}
return array;
}
Not a one-liner, but easy to read and fast.
add a comment |
The Collections.reverse
implies only to List
which is just one type of Collection
, you cannot cast a Queue
to a List
. But you can try casting it to a LinkedList
as:
Collections.reverse((LinkedList)queue);
Details:
I doubt that there is a built-in API for reversing the queue. You could still follow a conventional way of doing that using a Stack
as :
Stack<Integer> stack = new Stack<>();
while (!queue.isEmpty()) {
stack.add(queue.remove());
}
while (!stack.isEmpty()) {
queue.add(stack.pop());
}
and then convert to an array as you will
int res = queue.stream().mapToInt(Integer::intValue).toArray();
On the other hand, if a Deque
satisfies your needs currently, you can simply rely on the LinkedList
itself since it implements a Deque
as well. Then your current implementation would be as simple as :
LinkedList<Integer> dequeue = new LinkedList<>();
Collections.reverse(dequeue);
int res = dequeue.stream().mapToInt(Integer::intValue).toArray();
whether the queue is reversed is not important. An int array of the
reversed elements is what I need.
Another solution from what others have already suggested is to reverse the Stream
of the queue
and then mapToInt
to convert to an array as :
Queue<Integer> queue = new LinkedList<>();
int res = reverse(queue.stream()).mapToInt(Integer::intValue).toArray();
This uses a utility reverse
suggested by Stuart Marks in this answer such that:
@SuppressWarnings("unchecked")
static <T> Stream<T> reverse(Stream<T> input) {
Object temp = input.toArray();
return (Stream<T>) IntStream.range(0, temp.length)
.mapToObj(i -> temp[temp.length - i - 1]);
}
You should probably not be using theStack
class since it extendsVector
and is therefore synchronized, which is not needed here and only decreases performance.
– Marcono1234
yesterday
1
If using aDeque
it might be more efficient to useDeque.descendingIterator()
combined withSpliterators
andStreamSupport
, assuming only the reversed array is needed and not the reversedDeque
. The code will be more verbose, however.
– Slaw
yesterday
@Slaw It would be sure. Just that the intention of when I wrote the answer was to ensure the original store is reversed, but later the OP clarified that the reversed output is what matters eventually.
– nullpointer
yesterday
@Marcono1234 actualy, the JVM does away with thesynchronized
blocks within theVector
class is its most recent versions when it detects they're are not shared and their aquisition cost is negligible :) but yeah, on principle you shouldn't do that asVector
is not recommended to be used anymore.
– João Rebelo
18 hours ago
add a comment |
In Java8 version you can use Stream API to help you.
The skeleton of code like this:
int reversedQueue = queue.stream()
.collect(Collector.of(() -> new ArrayDeque<Integer>(), ArrayDeque::addFirst, (a,b)->a))
.stream().mapToInt(Integer::intValue).toArray();
It looks like your combiner ((a,b)->a
) is missingb
in the result
– Marcono1234
yesterday
@Marcono1234 There is no problem.The third parameter ofCollector.of
method is oneBinaryOperator
it's the combiner function for the new collector. In our code there only one collector,so can't miss any element in collector.
– TongChen
yesterday
it does not matter in this case probably since (if I understand it correctly) the combiner is only used for parallel streams. However, you should probably comment that the implementation is not correct, to prevent bugs in the future in case someone uses this code for a parallel stream. It should probably be(a, b) -> {b.addAll(a); return b;}
.
– Marcono1234
5 hours ago
add a comment |
Finally, I figure out this one line solution.
Integer intArray = queue.stream()
.collect(LinkedList::new, LinkedList::addFirst, LinkedList::addAll)
.toArray(new Integer[queue.size()]);
the int
version should like
int intArray = queue.stream()
.collect(LinkedList<Integer>::new, LinkedList::addFirst, LinkedList::addAll)
.stream()
.mapToInt(Integer::intValue)
.toArray();
Thanks @Hulk, add theint
version, but I think I like theInteger
version, simpler.
– Keijack
21 hours ago
add a comment |
You can use the LazyIterate
utility from Eclipse Collections as follows.
int res = LazyIterate.adapt(queue)
.collectInt(i -> i)
.toList()
.asReversed()
.toArray();
You can also use the Collectors2
class with a Java Stream.
int ints = queue.stream()
.collect(Collectors2.collectInt(i -> i, IntLists.mutable::empty))
.asReversed()
.toArray();
You can stream the int
values directly into a MutableIntList
, reverse it, and then convert it to an int
array.
int ints =
IntLists.mutable.ofAll(queue.stream().mapToInt(i -> i)).asReversed().toArray();
Finally, you can stream the int
values directly into a MutableIntStack
and convert it to an int
array.
int ints =
IntStacks.mutable.ofAll(queue.stream().mapToInt(i -> i)).toArray();
Note: I am a committer for Eclipse Collections.
add a comment |
This is one line, but it may not be very efficient:
int res = queue.stream()
.collect(LinkedList<Integer>::new, (l, e) -> l.addFirst(e), (l1, l2) -> l1.addAll(l2))
.stream()
.mapToInt(Integer::intValue)
.toArray();
If you want to be efficient and readable, you should continue using what you have now.
This does not reverse the queue (or its values)
– Marcono1234
yesterday
add a comment |
Here is a different solution using Stream
and Collections.reverse()
in one line of code:
Integer reversedArray = queue.stream()
.collect(Collectors.collectingAndThen(Collectors.toList(),
list -> {
Collections.reverse(list);
return list.toArray(new Integer[0]);
}
));
OR
int reversedArray = queue.stream()
.collect(Collectors.collectingAndThen(Collectors.toList(),
list -> {
Collections.reverse(list);
return list.stream()
.mapToInt(Integer::intValue)
.toArray();
}
));
add a comment |
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8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
First, please don't use raw types (do use the diamond operator). Not quite a one liner, but you could first convert to an int
and then use commons lang ArrayUtils.reverse(int)
like
Queue<Integer> queue = new LinkedList<>();
// ...
int arr = queue.stream().mapToInt(Integer::intValue).toArray();
ArrayUtils.reverse(arr);
You could also write your own int
reverse method that allowed for a fluent interface (e.g. return the int
) then you could make it a one liner. Like,
public static int reverse(int arr) {
for (int i = 0; i < arr.length / 2; i++) {
int temp = arr[i];
arr[i] = arr[arr.length - i - 1];
arr[arr.length - i - 1] = temp;
}
return arr;
}
And then
int arr = reverse(queue.stream().mapToInt(Integer::intValue).toArray());
but this wouldn't reverse the queue.
– nullpointer
yesterday
2
@nullpointer True. But, if the goal is a reversedint
then it isn't clear that the queue must also be reversed. In fact, I would assume the queue goes out of scope and theint
is returned to the caller.
– Elliott Frisch
yesterday
add a comment |
First, please don't use raw types (do use the diamond operator). Not quite a one liner, but you could first convert to an int
and then use commons lang ArrayUtils.reverse(int)
like
Queue<Integer> queue = new LinkedList<>();
// ...
int arr = queue.stream().mapToInt(Integer::intValue).toArray();
ArrayUtils.reverse(arr);
You could also write your own int
reverse method that allowed for a fluent interface (e.g. return the int
) then you could make it a one liner. Like,
public static int reverse(int arr) {
for (int i = 0; i < arr.length / 2; i++) {
int temp = arr[i];
arr[i] = arr[arr.length - i - 1];
arr[arr.length - i - 1] = temp;
}
return arr;
}
And then
int arr = reverse(queue.stream().mapToInt(Integer::intValue).toArray());
but this wouldn't reverse the queue.
– nullpointer
yesterday
2
@nullpointer True. But, if the goal is a reversedint
then it isn't clear that the queue must also be reversed. In fact, I would assume the queue goes out of scope and theint
is returned to the caller.
– Elliott Frisch
yesterday
add a comment |
First, please don't use raw types (do use the diamond operator). Not quite a one liner, but you could first convert to an int
and then use commons lang ArrayUtils.reverse(int)
like
Queue<Integer> queue = new LinkedList<>();
// ...
int arr = queue.stream().mapToInt(Integer::intValue).toArray();
ArrayUtils.reverse(arr);
You could also write your own int
reverse method that allowed for a fluent interface (e.g. return the int
) then you could make it a one liner. Like,
public static int reverse(int arr) {
for (int i = 0; i < arr.length / 2; i++) {
int temp = arr[i];
arr[i] = arr[arr.length - i - 1];
arr[arr.length - i - 1] = temp;
}
return arr;
}
And then
int arr = reverse(queue.stream().mapToInt(Integer::intValue).toArray());
First, please don't use raw types (do use the diamond operator). Not quite a one liner, but you could first convert to an int
and then use commons lang ArrayUtils.reverse(int)
like
Queue<Integer> queue = new LinkedList<>();
// ...
int arr = queue.stream().mapToInt(Integer::intValue).toArray();
ArrayUtils.reverse(arr);
You could also write your own int
reverse method that allowed for a fluent interface (e.g. return the int
) then you could make it a one liner. Like,
public static int reverse(int arr) {
for (int i = 0; i < arr.length / 2; i++) {
int temp = arr[i];
arr[i] = arr[arr.length - i - 1];
arr[arr.length - i - 1] = temp;
}
return arr;
}
And then
int arr = reverse(queue.stream().mapToInt(Integer::intValue).toArray());
answered yesterday
Elliott Frisch
153k1389178
153k1389178
but this wouldn't reverse the queue.
– nullpointer
yesterday
2
@nullpointer True. But, if the goal is a reversedint
then it isn't clear that the queue must also be reversed. In fact, I would assume the queue goes out of scope and theint
is returned to the caller.
– Elliott Frisch
yesterday
add a comment |
but this wouldn't reverse the queue.
– nullpointer
yesterday
2
@nullpointer True. But, if the goal is a reversedint
then it isn't clear that the queue must also be reversed. In fact, I would assume the queue goes out of scope and theint
is returned to the caller.
– Elliott Frisch
yesterday
but this wouldn't reverse the queue.
– nullpointer
yesterday
but this wouldn't reverse the queue.
– nullpointer
yesterday
2
2
@nullpointer True. But, if the goal is a reversed
int
then it isn't clear that the queue must also be reversed. In fact, I would assume the queue goes out of scope and the int
is returned to the caller.– Elliott Frisch
yesterday
@nullpointer True. But, if the goal is a reversed
int
then it isn't clear that the queue must also be reversed. In fact, I would assume the queue goes out of scope and the int
is returned to the caller.– Elliott Frisch
yesterday
add a comment |
No need to get fancy here.
static int toReversedArray(Queue<Integer> queue) {
int i = queue.size();
int array = new int[i];
for (int element : queue) {
array[--i] = element;
}
return array;
}
Not a one-liner, but easy to read and fast.
add a comment |
No need to get fancy here.
static int toReversedArray(Queue<Integer> queue) {
int i = queue.size();
int array = new int[i];
for (int element : queue) {
array[--i] = element;
}
return array;
}
Not a one-liner, but easy to read and fast.
add a comment |
No need to get fancy here.
static int toReversedArray(Queue<Integer> queue) {
int i = queue.size();
int array = new int[i];
for (int element : queue) {
array[--i] = element;
}
return array;
}
Not a one-liner, but easy to read and fast.
No need to get fancy here.
static int toReversedArray(Queue<Integer> queue) {
int i = queue.size();
int array = new int[i];
for (int element : queue) {
array[--i] = element;
}
return array;
}
Not a one-liner, but easy to read and fast.
answered 22 hours ago
xehpuk
4,3072335
4,3072335
add a comment |
add a comment |
The Collections.reverse
implies only to List
which is just one type of Collection
, you cannot cast a Queue
to a List
. But you can try casting it to a LinkedList
as:
Collections.reverse((LinkedList)queue);
Details:
I doubt that there is a built-in API for reversing the queue. You could still follow a conventional way of doing that using a Stack
as :
Stack<Integer> stack = new Stack<>();
while (!queue.isEmpty()) {
stack.add(queue.remove());
}
while (!stack.isEmpty()) {
queue.add(stack.pop());
}
and then convert to an array as you will
int res = queue.stream().mapToInt(Integer::intValue).toArray();
On the other hand, if a Deque
satisfies your needs currently, you can simply rely on the LinkedList
itself since it implements a Deque
as well. Then your current implementation would be as simple as :
LinkedList<Integer> dequeue = new LinkedList<>();
Collections.reverse(dequeue);
int res = dequeue.stream().mapToInt(Integer::intValue).toArray();
whether the queue is reversed is not important. An int array of the
reversed elements is what I need.
Another solution from what others have already suggested is to reverse the Stream
of the queue
and then mapToInt
to convert to an array as :
Queue<Integer> queue = new LinkedList<>();
int res = reverse(queue.stream()).mapToInt(Integer::intValue).toArray();
This uses a utility reverse
suggested by Stuart Marks in this answer such that:
@SuppressWarnings("unchecked")
static <T> Stream<T> reverse(Stream<T> input) {
Object temp = input.toArray();
return (Stream<T>) IntStream.range(0, temp.length)
.mapToObj(i -> temp[temp.length - i - 1]);
}
You should probably not be using theStack
class since it extendsVector
and is therefore synchronized, which is not needed here and only decreases performance.
– Marcono1234
yesterday
1
If using aDeque
it might be more efficient to useDeque.descendingIterator()
combined withSpliterators
andStreamSupport
, assuming only the reversed array is needed and not the reversedDeque
. The code will be more verbose, however.
– Slaw
yesterday
@Slaw It would be sure. Just that the intention of when I wrote the answer was to ensure the original store is reversed, but later the OP clarified that the reversed output is what matters eventually.
– nullpointer
yesterday
@Marcono1234 actualy, the JVM does away with thesynchronized
blocks within theVector
class is its most recent versions when it detects they're are not shared and their aquisition cost is negligible :) but yeah, on principle you shouldn't do that asVector
is not recommended to be used anymore.
– João Rebelo
18 hours ago
add a comment |
The Collections.reverse
implies only to List
which is just one type of Collection
, you cannot cast a Queue
to a List
. But you can try casting it to a LinkedList
as:
Collections.reverse((LinkedList)queue);
Details:
I doubt that there is a built-in API for reversing the queue. You could still follow a conventional way of doing that using a Stack
as :
Stack<Integer> stack = new Stack<>();
while (!queue.isEmpty()) {
stack.add(queue.remove());
}
while (!stack.isEmpty()) {
queue.add(stack.pop());
}
and then convert to an array as you will
int res = queue.stream().mapToInt(Integer::intValue).toArray();
On the other hand, if a Deque
satisfies your needs currently, you can simply rely on the LinkedList
itself since it implements a Deque
as well. Then your current implementation would be as simple as :
LinkedList<Integer> dequeue = new LinkedList<>();
Collections.reverse(dequeue);
int res = dequeue.stream().mapToInt(Integer::intValue).toArray();
whether the queue is reversed is not important. An int array of the
reversed elements is what I need.
Another solution from what others have already suggested is to reverse the Stream
of the queue
and then mapToInt
to convert to an array as :
Queue<Integer> queue = new LinkedList<>();
int res = reverse(queue.stream()).mapToInt(Integer::intValue).toArray();
This uses a utility reverse
suggested by Stuart Marks in this answer such that:
@SuppressWarnings("unchecked")
static <T> Stream<T> reverse(Stream<T> input) {
Object temp = input.toArray();
return (Stream<T>) IntStream.range(0, temp.length)
.mapToObj(i -> temp[temp.length - i - 1]);
}
You should probably not be using theStack
class since it extendsVector
and is therefore synchronized, which is not needed here and only decreases performance.
– Marcono1234
yesterday
1
If using aDeque
it might be more efficient to useDeque.descendingIterator()
combined withSpliterators
andStreamSupport
, assuming only the reversed array is needed and not the reversedDeque
. The code will be more verbose, however.
– Slaw
yesterday
@Slaw It would be sure. Just that the intention of when I wrote the answer was to ensure the original store is reversed, but later the OP clarified that the reversed output is what matters eventually.
– nullpointer
yesterday
@Marcono1234 actualy, the JVM does away with thesynchronized
blocks within theVector
class is its most recent versions when it detects they're are not shared and their aquisition cost is negligible :) but yeah, on principle you shouldn't do that asVector
is not recommended to be used anymore.
– João Rebelo
18 hours ago
add a comment |
The Collections.reverse
implies only to List
which is just one type of Collection
, you cannot cast a Queue
to a List
. But you can try casting it to a LinkedList
as:
Collections.reverse((LinkedList)queue);
Details:
I doubt that there is a built-in API for reversing the queue. You could still follow a conventional way of doing that using a Stack
as :
Stack<Integer> stack = new Stack<>();
while (!queue.isEmpty()) {
stack.add(queue.remove());
}
while (!stack.isEmpty()) {
queue.add(stack.pop());
}
and then convert to an array as you will
int res = queue.stream().mapToInt(Integer::intValue).toArray();
On the other hand, if a Deque
satisfies your needs currently, you can simply rely on the LinkedList
itself since it implements a Deque
as well. Then your current implementation would be as simple as :
LinkedList<Integer> dequeue = new LinkedList<>();
Collections.reverse(dequeue);
int res = dequeue.stream().mapToInt(Integer::intValue).toArray();
whether the queue is reversed is not important. An int array of the
reversed elements is what I need.
Another solution from what others have already suggested is to reverse the Stream
of the queue
and then mapToInt
to convert to an array as :
Queue<Integer> queue = new LinkedList<>();
int res = reverse(queue.stream()).mapToInt(Integer::intValue).toArray();
This uses a utility reverse
suggested by Stuart Marks in this answer such that:
@SuppressWarnings("unchecked")
static <T> Stream<T> reverse(Stream<T> input) {
Object temp = input.toArray();
return (Stream<T>) IntStream.range(0, temp.length)
.mapToObj(i -> temp[temp.length - i - 1]);
}
The Collections.reverse
implies only to List
which is just one type of Collection
, you cannot cast a Queue
to a List
. But you can try casting it to a LinkedList
as:
Collections.reverse((LinkedList)queue);
Details:
I doubt that there is a built-in API for reversing the queue. You could still follow a conventional way of doing that using a Stack
as :
Stack<Integer> stack = new Stack<>();
while (!queue.isEmpty()) {
stack.add(queue.remove());
}
while (!stack.isEmpty()) {
queue.add(stack.pop());
}
and then convert to an array as you will
int res = queue.stream().mapToInt(Integer::intValue).toArray();
On the other hand, if a Deque
satisfies your needs currently, you can simply rely on the LinkedList
itself since it implements a Deque
as well. Then your current implementation would be as simple as :
LinkedList<Integer> dequeue = new LinkedList<>();
Collections.reverse(dequeue);
int res = dequeue.stream().mapToInt(Integer::intValue).toArray();
whether the queue is reversed is not important. An int array of the
reversed elements is what I need.
Another solution from what others have already suggested is to reverse the Stream
of the queue
and then mapToInt
to convert to an array as :
Queue<Integer> queue = new LinkedList<>();
int res = reverse(queue.stream()).mapToInt(Integer::intValue).toArray();
This uses a utility reverse
suggested by Stuart Marks in this answer such that:
@SuppressWarnings("unchecked")
static <T> Stream<T> reverse(Stream<T> input) {
Object temp = input.toArray();
return (Stream<T>) IntStream.range(0, temp.length)
.mapToObj(i -> temp[temp.length - i - 1]);
}
edited 20 hours ago
answered yesterday
nullpointer
43.2k1093178
43.2k1093178
You should probably not be using theStack
class since it extendsVector
and is therefore synchronized, which is not needed here and only decreases performance.
– Marcono1234
yesterday
1
If using aDeque
it might be more efficient to useDeque.descendingIterator()
combined withSpliterators
andStreamSupport
, assuming only the reversed array is needed and not the reversedDeque
. The code will be more verbose, however.
– Slaw
yesterday
@Slaw It would be sure. Just that the intention of when I wrote the answer was to ensure the original store is reversed, but later the OP clarified that the reversed output is what matters eventually.
– nullpointer
yesterday
@Marcono1234 actualy, the JVM does away with thesynchronized
blocks within theVector
class is its most recent versions when it detects they're are not shared and their aquisition cost is negligible :) but yeah, on principle you shouldn't do that asVector
is not recommended to be used anymore.
– João Rebelo
18 hours ago
add a comment |
You should probably not be using theStack
class since it extendsVector
and is therefore synchronized, which is not needed here and only decreases performance.
– Marcono1234
yesterday
1
If using aDeque
it might be more efficient to useDeque.descendingIterator()
combined withSpliterators
andStreamSupport
, assuming only the reversed array is needed and not the reversedDeque
. The code will be more verbose, however.
– Slaw
yesterday
@Slaw It would be sure. Just that the intention of when I wrote the answer was to ensure the original store is reversed, but later the OP clarified that the reversed output is what matters eventually.
– nullpointer
yesterday
@Marcono1234 actualy, the JVM does away with thesynchronized
blocks within theVector
class is its most recent versions when it detects they're are not shared and their aquisition cost is negligible :) but yeah, on principle you shouldn't do that asVector
is not recommended to be used anymore.
– João Rebelo
18 hours ago
You should probably not be using the
Stack
class since it extends Vector
and is therefore synchronized, which is not needed here and only decreases performance.– Marcono1234
yesterday
You should probably not be using the
Stack
class since it extends Vector
and is therefore synchronized, which is not needed here and only decreases performance.– Marcono1234
yesterday
1
1
If using a
Deque
it might be more efficient to use Deque.descendingIterator()
combined with Spliterators
and StreamSupport
, assuming only the reversed array is needed and not the reversed Deque
. The code will be more verbose, however.– Slaw
yesterday
If using a
Deque
it might be more efficient to use Deque.descendingIterator()
combined with Spliterators
and StreamSupport
, assuming only the reversed array is needed and not the reversed Deque
. The code will be more verbose, however.– Slaw
yesterday
@Slaw It would be sure. Just that the intention of when I wrote the answer was to ensure the original store is reversed, but later the OP clarified that the reversed output is what matters eventually.
– nullpointer
yesterday
@Slaw It would be sure. Just that the intention of when I wrote the answer was to ensure the original store is reversed, but later the OP clarified that the reversed output is what matters eventually.
– nullpointer
yesterday
@Marcono1234 actualy, the JVM does away with the
synchronized
blocks within the Vector
class is its most recent versions when it detects they're are not shared and their aquisition cost is negligible :) but yeah, on principle you shouldn't do that as Vector
is not recommended to be used anymore.– João Rebelo
18 hours ago
@Marcono1234 actualy, the JVM does away with the
synchronized
blocks within the Vector
class is its most recent versions when it detects they're are not shared and their aquisition cost is negligible :) but yeah, on principle you shouldn't do that as Vector
is not recommended to be used anymore.– João Rebelo
18 hours ago
add a comment |
In Java8 version you can use Stream API to help you.
The skeleton of code like this:
int reversedQueue = queue.stream()
.collect(Collector.of(() -> new ArrayDeque<Integer>(), ArrayDeque::addFirst, (a,b)->a))
.stream().mapToInt(Integer::intValue).toArray();
It looks like your combiner ((a,b)->a
) is missingb
in the result
– Marcono1234
yesterday
@Marcono1234 There is no problem.The third parameter ofCollector.of
method is oneBinaryOperator
it's the combiner function for the new collector. In our code there only one collector,so can't miss any element in collector.
– TongChen
yesterday
it does not matter in this case probably since (if I understand it correctly) the combiner is only used for parallel streams. However, you should probably comment that the implementation is not correct, to prevent bugs in the future in case someone uses this code for a parallel stream. It should probably be(a, b) -> {b.addAll(a); return b;}
.
– Marcono1234
5 hours ago
add a comment |
In Java8 version you can use Stream API to help you.
The skeleton of code like this:
int reversedQueue = queue.stream()
.collect(Collector.of(() -> new ArrayDeque<Integer>(), ArrayDeque::addFirst, (a,b)->a))
.stream().mapToInt(Integer::intValue).toArray();
It looks like your combiner ((a,b)->a
) is missingb
in the result
– Marcono1234
yesterday
@Marcono1234 There is no problem.The third parameter ofCollector.of
method is oneBinaryOperator
it's the combiner function for the new collector. In our code there only one collector,so can't miss any element in collector.
– TongChen
yesterday
it does not matter in this case probably since (if I understand it correctly) the combiner is only used for parallel streams. However, you should probably comment that the implementation is not correct, to prevent bugs in the future in case someone uses this code for a parallel stream. It should probably be(a, b) -> {b.addAll(a); return b;}
.
– Marcono1234
5 hours ago
add a comment |
In Java8 version you can use Stream API to help you.
The skeleton of code like this:
int reversedQueue = queue.stream()
.collect(Collector.of(() -> new ArrayDeque<Integer>(), ArrayDeque::addFirst, (a,b)->a))
.stream().mapToInt(Integer::intValue).toArray();
In Java8 version you can use Stream API to help you.
The skeleton of code like this:
int reversedQueue = queue.stream()
.collect(Collector.of(() -> new ArrayDeque<Integer>(), ArrayDeque::addFirst, (a,b)->a))
.stream().mapToInt(Integer::intValue).toArray();
answered yesterday
TongChen
1958
1958
It looks like your combiner ((a,b)->a
) is missingb
in the result
– Marcono1234
yesterday
@Marcono1234 There is no problem.The third parameter ofCollector.of
method is oneBinaryOperator
it's the combiner function for the new collector. In our code there only one collector,so can't miss any element in collector.
– TongChen
yesterday
it does not matter in this case probably since (if I understand it correctly) the combiner is only used for parallel streams. However, you should probably comment that the implementation is not correct, to prevent bugs in the future in case someone uses this code for a parallel stream. It should probably be(a, b) -> {b.addAll(a); return b;}
.
– Marcono1234
5 hours ago
add a comment |
It looks like your combiner ((a,b)->a
) is missingb
in the result
– Marcono1234
yesterday
@Marcono1234 There is no problem.The third parameter ofCollector.of
method is oneBinaryOperator
it's the combiner function for the new collector. In our code there only one collector,so can't miss any element in collector.
– TongChen
yesterday
it does not matter in this case probably since (if I understand it correctly) the combiner is only used for parallel streams. However, you should probably comment that the implementation is not correct, to prevent bugs in the future in case someone uses this code for a parallel stream. It should probably be(a, b) -> {b.addAll(a); return b;}
.
– Marcono1234
5 hours ago
It looks like your combiner (
(a,b)->a
) is missing b
in the result– Marcono1234
yesterday
It looks like your combiner (
(a,b)->a
) is missing b
in the result– Marcono1234
yesterday
@Marcono1234 There is no problem.The third parameter of
Collector.of
method is one BinaryOperator
it's the combiner function for the new collector. In our code there only one collector,so can't miss any element in collector.– TongChen
yesterday
@Marcono1234 There is no problem.The third parameter of
Collector.of
method is one BinaryOperator
it's the combiner function for the new collector. In our code there only one collector,so can't miss any element in collector.– TongChen
yesterday
it does not matter in this case probably since (if I understand it correctly) the combiner is only used for parallel streams. However, you should probably comment that the implementation is not correct, to prevent bugs in the future in case someone uses this code for a parallel stream. It should probably be
(a, b) -> {b.addAll(a); return b;}
.– Marcono1234
5 hours ago
it does not matter in this case probably since (if I understand it correctly) the combiner is only used for parallel streams. However, you should probably comment that the implementation is not correct, to prevent bugs in the future in case someone uses this code for a parallel stream. It should probably be
(a, b) -> {b.addAll(a); return b;}
.– Marcono1234
5 hours ago
add a comment |
Finally, I figure out this one line solution.
Integer intArray = queue.stream()
.collect(LinkedList::new, LinkedList::addFirst, LinkedList::addAll)
.toArray(new Integer[queue.size()]);
the int
version should like
int intArray = queue.stream()
.collect(LinkedList<Integer>::new, LinkedList::addFirst, LinkedList::addAll)
.stream()
.mapToInt(Integer::intValue)
.toArray();
Thanks @Hulk, add theint
version, but I think I like theInteger
version, simpler.
– Keijack
21 hours ago
add a comment |
Finally, I figure out this one line solution.
Integer intArray = queue.stream()
.collect(LinkedList::new, LinkedList::addFirst, LinkedList::addAll)
.toArray(new Integer[queue.size()]);
the int
version should like
int intArray = queue.stream()
.collect(LinkedList<Integer>::new, LinkedList::addFirst, LinkedList::addAll)
.stream()
.mapToInt(Integer::intValue)
.toArray();
Thanks @Hulk, add theint
version, but I think I like theInteger
version, simpler.
– Keijack
21 hours ago
add a comment |
Finally, I figure out this one line solution.
Integer intArray = queue.stream()
.collect(LinkedList::new, LinkedList::addFirst, LinkedList::addAll)
.toArray(new Integer[queue.size()]);
the int
version should like
int intArray = queue.stream()
.collect(LinkedList<Integer>::new, LinkedList::addFirst, LinkedList::addAll)
.stream()
.mapToInt(Integer::intValue)
.toArray();
Finally, I figure out this one line solution.
Integer intArray = queue.stream()
.collect(LinkedList::new, LinkedList::addFirst, LinkedList::addAll)
.toArray(new Integer[queue.size()]);
the int
version should like
int intArray = queue.stream()
.collect(LinkedList<Integer>::new, LinkedList::addFirst, LinkedList::addAll)
.stream()
.mapToInt(Integer::intValue)
.toArray();
edited 21 hours ago
answered yesterday
Keijack
1666
1666
Thanks @Hulk, add theint
version, but I think I like theInteger
version, simpler.
– Keijack
21 hours ago
add a comment |
Thanks @Hulk, add theint
version, but I think I like theInteger
version, simpler.
– Keijack
21 hours ago
Thanks @Hulk, add the
int
version, but I think I like the Integer
version, simpler.– Keijack
21 hours ago
Thanks @Hulk, add the
int
version, but I think I like the Integer
version, simpler.– Keijack
21 hours ago
add a comment |
You can use the LazyIterate
utility from Eclipse Collections as follows.
int res = LazyIterate.adapt(queue)
.collectInt(i -> i)
.toList()
.asReversed()
.toArray();
You can also use the Collectors2
class with a Java Stream.
int ints = queue.stream()
.collect(Collectors2.collectInt(i -> i, IntLists.mutable::empty))
.asReversed()
.toArray();
You can stream the int
values directly into a MutableIntList
, reverse it, and then convert it to an int
array.
int ints =
IntLists.mutable.ofAll(queue.stream().mapToInt(i -> i)).asReversed().toArray();
Finally, you can stream the int
values directly into a MutableIntStack
and convert it to an int
array.
int ints =
IntStacks.mutable.ofAll(queue.stream().mapToInt(i -> i)).toArray();
Note: I am a committer for Eclipse Collections.
add a comment |
You can use the LazyIterate
utility from Eclipse Collections as follows.
int res = LazyIterate.adapt(queue)
.collectInt(i -> i)
.toList()
.asReversed()
.toArray();
You can also use the Collectors2
class with a Java Stream.
int ints = queue.stream()
.collect(Collectors2.collectInt(i -> i, IntLists.mutable::empty))
.asReversed()
.toArray();
You can stream the int
values directly into a MutableIntList
, reverse it, and then convert it to an int
array.
int ints =
IntLists.mutable.ofAll(queue.stream().mapToInt(i -> i)).asReversed().toArray();
Finally, you can stream the int
values directly into a MutableIntStack
and convert it to an int
array.
int ints =
IntStacks.mutable.ofAll(queue.stream().mapToInt(i -> i)).toArray();
Note: I am a committer for Eclipse Collections.
add a comment |
You can use the LazyIterate
utility from Eclipse Collections as follows.
int res = LazyIterate.adapt(queue)
.collectInt(i -> i)
.toList()
.asReversed()
.toArray();
You can also use the Collectors2
class with a Java Stream.
int ints = queue.stream()
.collect(Collectors2.collectInt(i -> i, IntLists.mutable::empty))
.asReversed()
.toArray();
You can stream the int
values directly into a MutableIntList
, reverse it, and then convert it to an int
array.
int ints =
IntLists.mutable.ofAll(queue.stream().mapToInt(i -> i)).asReversed().toArray();
Finally, you can stream the int
values directly into a MutableIntStack
and convert it to an int
array.
int ints =
IntStacks.mutable.ofAll(queue.stream().mapToInt(i -> i)).toArray();
Note: I am a committer for Eclipse Collections.
You can use the LazyIterate
utility from Eclipse Collections as follows.
int res = LazyIterate.adapt(queue)
.collectInt(i -> i)
.toList()
.asReversed()
.toArray();
You can also use the Collectors2
class with a Java Stream.
int ints = queue.stream()
.collect(Collectors2.collectInt(i -> i, IntLists.mutable::empty))
.asReversed()
.toArray();
You can stream the int
values directly into a MutableIntList
, reverse it, and then convert it to an int
array.
int ints =
IntLists.mutable.ofAll(queue.stream().mapToInt(i -> i)).asReversed().toArray();
Finally, you can stream the int
values directly into a MutableIntStack
and convert it to an int
array.
int ints =
IntStacks.mutable.ofAll(queue.stream().mapToInt(i -> i)).toArray();
Note: I am a committer for Eclipse Collections.
edited 22 hours ago
answered yesterday
Donald Raab
4,21112029
4,21112029
add a comment |
add a comment |
This is one line, but it may not be very efficient:
int res = queue.stream()
.collect(LinkedList<Integer>::new, (l, e) -> l.addFirst(e), (l1, l2) -> l1.addAll(l2))
.stream()
.mapToInt(Integer::intValue)
.toArray();
If you want to be efficient and readable, you should continue using what you have now.
This does not reverse the queue (or its values)
– Marcono1234
yesterday
add a comment |
This is one line, but it may not be very efficient:
int res = queue.stream()
.collect(LinkedList<Integer>::new, (l, e) -> l.addFirst(e), (l1, l2) -> l1.addAll(l2))
.stream()
.mapToInt(Integer::intValue)
.toArray();
If you want to be efficient and readable, you should continue using what you have now.
This does not reverse the queue (or its values)
– Marcono1234
yesterday
add a comment |
This is one line, but it may not be very efficient:
int res = queue.stream()
.collect(LinkedList<Integer>::new, (l, e) -> l.addFirst(e), (l1, l2) -> l1.addAll(l2))
.stream()
.mapToInt(Integer::intValue)
.toArray();
If you want to be efficient and readable, you should continue using what you have now.
This is one line, but it may not be very efficient:
int res = queue.stream()
.collect(LinkedList<Integer>::new, (l, e) -> l.addFirst(e), (l1, l2) -> l1.addAll(l2))
.stream()
.mapToInt(Integer::intValue)
.toArray();
If you want to be efficient and readable, you should continue using what you have now.
edited 21 hours ago
ZhaoGang
1,6181015
1,6181015
answered yesterday
Jai
5,73311231
5,73311231
This does not reverse the queue (or its values)
– Marcono1234
yesterday
add a comment |
This does not reverse the queue (or its values)
– Marcono1234
yesterday
This does not reverse the queue (or its values)
– Marcono1234
yesterday
This does not reverse the queue (or its values)
– Marcono1234
yesterday
add a comment |
Here is a different solution using Stream
and Collections.reverse()
in one line of code:
Integer reversedArray = queue.stream()
.collect(Collectors.collectingAndThen(Collectors.toList(),
list -> {
Collections.reverse(list);
return list.toArray(new Integer[0]);
}
));
OR
int reversedArray = queue.stream()
.collect(Collectors.collectingAndThen(Collectors.toList(),
list -> {
Collections.reverse(list);
return list.stream()
.mapToInt(Integer::intValue)
.toArray();
}
));
add a comment |
Here is a different solution using Stream
and Collections.reverse()
in one line of code:
Integer reversedArray = queue.stream()
.collect(Collectors.collectingAndThen(Collectors.toList(),
list -> {
Collections.reverse(list);
return list.toArray(new Integer[0]);
}
));
OR
int reversedArray = queue.stream()
.collect(Collectors.collectingAndThen(Collectors.toList(),
list -> {
Collections.reverse(list);
return list.stream()
.mapToInt(Integer::intValue)
.toArray();
}
));
add a comment |
Here is a different solution using Stream
and Collections.reverse()
in one line of code:
Integer reversedArray = queue.stream()
.collect(Collectors.collectingAndThen(Collectors.toList(),
list -> {
Collections.reverse(list);
return list.toArray(new Integer[0]);
}
));
OR
int reversedArray = queue.stream()
.collect(Collectors.collectingAndThen(Collectors.toList(),
list -> {
Collections.reverse(list);
return list.stream()
.mapToInt(Integer::intValue)
.toArray();
}
));
Here is a different solution using Stream
and Collections.reverse()
in one line of code:
Integer reversedArray = queue.stream()
.collect(Collectors.collectingAndThen(Collectors.toList(),
list -> {
Collections.reverse(list);
return list.toArray(new Integer[0]);
}
));
OR
int reversedArray = queue.stream()
.collect(Collectors.collectingAndThen(Collectors.toList(),
list -> {
Collections.reverse(list);
return list.stream()
.mapToInt(Integer::intValue)
.toArray();
}
));
edited 19 hours ago
answered 20 hours ago
aminography
5,51821130
5,51821130
add a comment |
add a comment |
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