Mixing
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
edited Jan 4 at 9:33
Moira
5,52221937
asked Jan 4 at 1:18
ZhaoGangZhaoGang
1,9011116
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
edited Jan 4 at 9:33
Moira
5,52221937
asked Jan 4 at 1:18
ZhaoGangZhaoGang
1,9011116
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
edited Jan 4 at 9:33
Moira
5,52221937
asked Jan 4 at 1:18
ZhaoGangZhaoGang
1,9011116
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 Jan 4 at 9:33
Moira
5,52221937
asked Jan 4 at 1:18
ZhaoGangZhaoGang
1,9011116
edited Jan 4 at 9:33
Moira
5,52221937
asked Jan 4 at 1:18
ZhaoGangZhaoGang
1,9011116
edited Jan 4 at 9:33
Moira
5,52221937
edited Jan 4 at 9:33
Moira
5,52221937
edited Jan 4 at 9:33
Moira
5,52221937
5,52221937
asked Jan 4 at 1:18
ZhaoGangZhaoGang
1,9011116
asked Jan 4 at 1:18
ZhaoGangZhaoGang
1,9011116
asked Jan 4 at 1:18
ZhaoGangZhaoGang
1,9011116
1,9011116
add a comment |
add a comment |
9 Answers
9
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());
answered Jan 4 at 1:35
Elliott FrischElliott Frisch
153k1389178
but this wouldn't reverse the queue.
– nullpointer
Jan 4 at 1:39
2
@nullpointer True. But, if the goal is a reversedintthen it isn't clear that the queue must also be reversed. In fact, I would assume the queue goes out of scope and theintis returned to the caller.
– Elliott Frisch
Jan 4 at 1:42
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.
answered Jan 4 at 7:02
xehpukxehpuk
4,4322335
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]);
}
answered Jan 4 at 1:31
nullpointernullpointer
46k1198188
You should probably not be using theStackclass since it extendsVectorand is therefore synchronized, which is not needed here and only decreases performance.
– Marcono1234
Jan 4 at 2:39
2
If using aDequeit might be more efficient to useDeque.descendingIterator()combined withSpliteratorsandStreamSupport, assuming only the reversed array is needed and not the reversedDeque. The code will be more verbose, however.
– Slaw
Jan 4 at 2:39
@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
Jan 4 at 3:43
@Marcono1234 actualy, the JVM does away with thesynchronizedblocks within theVectorclass 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 asVectoris not recommended to be used anymore.
– João Rebelo
Jan 4 at 10:43
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();
answered Jan 4 at 1:59
TongChenTongChen
27529
It looks like your combiner ((a,b)->a) is missingbin the result
– Marcono1234
Jan 4 at 2:52
@Marcono1234 There is no problem.The third parameter ofCollector.ofmethod is oneBinaryOperatorit's the combiner function for the new collector. In our code there only one collector,so can't miss any element in collector.
– TongChen
Jan 4 at 3:24
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
Jan 5 at 0:00
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.
answered Jan 4 at 3:36
Donald RaabDonald Raab
4,22112029
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();
answered Jan 4 at 1:42
KeijackKeijack
1666
Thanks @Hulk, add theintversion, but I think I like theIntegerversion, simpler.
– Keijack
Jan 4 at 8:05
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.
edited Jan 4 at 8:01
ZhaoGang
1,9011116
answered Jan 4 at 1:53
JaiJai
5,76811231
This does not reverse the queue (or its values)
– Marcono1234
Jan 4 at 2:50
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();
}
));
answered Jan 4 at 9:17
aminographyaminography
5,91321131
add a comment |
Here's a way that creates a reversed array without reversing the queue:
int i = { queue.size() };
int array = new int[i[0]];
queue.forEach(n -> array[--i[0]] = n);
The above code is quite hacky due to the impossibility to modify local variables from within lambda expressions. So here i needs to be a one-element array, to overcome this restriction.
Note: bear in mind that I've come to this solution just for fun :)
answered Jan 5 at 16:53
Federico Peralta SchaffnerFederico Peralta Schaffner
22.3k43573
add a comment |
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9 Answers
9
active
oldest
votes
9 Answers
9
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());
answered Jan 4 at 1:35
Elliott FrischElliott Frisch
153k1389178
but this wouldn't reverse the queue.
– nullpointer
Jan 4 at 1:39
2
@nullpointer True. But, if the goal is a reversedintthen it isn't clear that the queue must also be reversed. In fact, I would assume the queue goes out of scope and theintis returned to the caller.
– Elliott Frisch
Jan 4 at 1:42
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());
answered Jan 4 at 1:35
Elliott FrischElliott Frisch
153k1389178
but this wouldn't reverse the queue.
– nullpointer
Jan 4 at 1:39
2
@nullpointer True. But, if the goal is a reversedintthen it isn't clear that the queue must also be reversed. In fact, I would assume the queue goes out of scope and theintis returned to the caller.
– Elliott Frisch
Jan 4 at 1:42
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());
answered Jan 4 at 1:35
Elliott FrischElliott Frisch
153k1389178
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 Jan 4 at 1:35
Elliott FrischElliott Frisch
153k1389178
answered Jan 4 at 1:35
Elliott FrischElliott Frisch
153k1389178
answered Jan 4 at 1:35
Elliott FrischElliott Frisch
153k1389178
answered Jan 4 at 1:35
Elliott FrischElliott Frisch
153k1389178
153k1389178
but this wouldn't reverse the queue.
– nullpointer
Jan 4 at 1:39
2
@nullpointer True. But, if the goal is a reversedintthen it isn't clear that the queue must also be reversed. In fact, I would assume the queue goes out of scope and theintis returned to the caller.
– Elliott Frisch
Jan 4 at 1:42
add a comment |
but this wouldn't reverse the queue.
– nullpointer
Jan 4 at 1:39
2
@nullpointer True. But, if the goal is a reversedintthen it isn't clear that the queue must also be reversed. In fact, I would assume the queue goes out of scope and theintis returned to the caller.
– Elliott Frisch
Jan 4 at 1:42
but this wouldn't reverse the queue.
– nullpointer
Jan 4 at 1:39
but this wouldn't reverse the queue.
– nullpointer
Jan 4 at 1:39
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
Jan 4 at 1:42
@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
Jan 4 at 1:42
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.
answered Jan 4 at 7:02
xehpukxehpuk
4,4322335
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.
answered Jan 4 at 7:02
xehpukxehpuk
4,4322335
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.
answered Jan 4 at 7:02
xehpukxehpuk
4,4322335
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 Jan 4 at 7:02
xehpukxehpuk
4,4322335
answered Jan 4 at 7:02
xehpukxehpuk
4,4322335
answered Jan 4 at 7:02
xehpukxehpuk
4,4322335
answered Jan 4 at 7:02
xehpukxehpuk
4,4322335
4,4322335
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]);
}
answered Jan 4 at 1:31
nullpointernullpointer
46k1198188
You should probably not be using theStackclass since it extendsVectorand is therefore synchronized, which is not needed here and only decreases performance.
– Marcono1234
Jan 4 at 2:39
2
If using aDequeit might be more efficient to useDeque.descendingIterator()combined withSpliteratorsandStreamSupport, assuming only the reversed array is needed and not the reversedDeque. The code will be more verbose, however.
– Slaw
Jan 4 at 2:39
@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
Jan 4 at 3:43
@Marcono1234 actualy, the JVM does away with thesynchronizedblocks within theVectorclass 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 asVectoris not recommended to be used anymore.
– João Rebelo
Jan 4 at 10:43
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]);
}
answered Jan 4 at 1:31
nullpointernullpointer
46k1198188
You should probably not be using theStackclass since it extendsVectorand is therefore synchronized, which is not needed here and only decreases performance.
– Marcono1234
Jan 4 at 2:39
2
If using aDequeit might be more efficient to useDeque.descendingIterator()combined withSpliteratorsandStreamSupport, assuming only the reversed array is needed and not the reversedDeque. The code will be more verbose, however.
– Slaw
Jan 4 at 2:39
@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
Jan 4 at 3:43
@Marcono1234 actualy, the JVM does away with thesynchronizedblocks within theVectorclass 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 asVectoris not recommended to be used anymore.
– João Rebelo
Jan 4 at 10:43
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]);
}
answered Jan 4 at 1:31
nullpointernullpointer
46k1198188
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]);
}
answered Jan 4 at 1:31
nullpointernullpointer
46k1198188
edited Jan 4 at 8:31
answered Jan 4 at 1:31
nullpointernullpointer
46k1198188
answered Jan 4 at 1:31
nullpointernullpointer
46k1198188
answered Jan 4 at 1:31
nullpointernullpointer
46k1198188
46k1198188
You should probably not be using theStackclass since it extendsVectorand is therefore synchronized, which is not needed here and only decreases performance.
– Marcono1234
Jan 4 at 2:39
2
If using aDequeit might be more efficient to useDeque.descendingIterator()combined withSpliteratorsandStreamSupport, assuming only the reversed array is needed and not the reversedDeque. The code will be more verbose, however.
– Slaw
Jan 4 at 2:39
@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
Jan 4 at 3:43
@Marcono1234 actualy, the JVM does away with thesynchronizedblocks within theVectorclass 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 asVectoris not recommended to be used anymore.
– João Rebelo
Jan 4 at 10:43
add a comment |
You should probably not be using theStackclass since it extendsVectorand is therefore synchronized, which is not needed here and only decreases performance.
– Marcono1234
Jan 4 at 2:39
2
If using aDequeit might be more efficient to useDeque.descendingIterator()combined withSpliteratorsandStreamSupport, assuming only the reversed array is needed and not the reversedDeque. The code will be more verbose, however.
– Slaw
Jan 4 at 2:39
@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
Jan 4 at 3:43
@Marcono1234 actualy, the JVM does away with thesynchronizedblocks within theVectorclass 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 asVectoris not recommended to be used anymore.
– João Rebelo
Jan 4 at 10:43
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
Jan 4 at 2:39
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
Jan 4 at 2:39
2
2
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
Jan 4 at 2:39
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
Jan 4 at 2:39
@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
Jan 4 at 3:43
@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
Jan 4 at 3:43
@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
Jan 4 at 10:43
@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
Jan 4 at 10:43
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();
answered Jan 4 at 1:59
TongChenTongChen
27529
It looks like your combiner ((a,b)->a) is missingbin the result
– Marcono1234
Jan 4 at 2:52
@Marcono1234 There is no problem.The third parameter ofCollector.ofmethod is oneBinaryOperatorit's the combiner function for the new collector. In our code there only one collector,so can't miss any element in collector.
– TongChen
Jan 4 at 3:24
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
Jan 5 at 0:00
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();
answered Jan 4 at 1:59
TongChenTongChen
27529
It looks like your combiner ((a,b)->a) is missingbin the result
– Marcono1234
Jan 4 at 2:52
@Marcono1234 There is no problem.The third parameter ofCollector.ofmethod is oneBinaryOperatorit's the combiner function for the new collector. In our code there only one collector,so can't miss any element in collector.
– TongChen
Jan 4 at 3:24
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
Jan 5 at 0:00
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();
answered Jan 4 at 1:59
TongChenTongChen
27529
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 Jan 4 at 1:59
TongChenTongChen
27529
answered Jan 4 at 1:59
TongChenTongChen
27529
answered Jan 4 at 1:59
TongChenTongChen
27529
answered Jan 4 at 1:59
TongChenTongChen
27529
27529
It looks like your combiner ((a,b)->a) is missingbin the result
– Marcono1234
Jan 4 at 2:52
@Marcono1234 There is no problem.The third parameter ofCollector.ofmethod is oneBinaryOperatorit's the combiner function for the new collector. In our code there only one collector,so can't miss any element in collector.
– TongChen
Jan 4 at 3:24
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
Jan 5 at 0:00
add a comment |
It looks like your combiner ((a,b)->a) is missingbin the result
– Marcono1234
Jan 4 at 2:52
@Marcono1234 There is no problem.The third parameter ofCollector.ofmethod is oneBinaryOperatorit's the combiner function for the new collector. In our code there only one collector,so can't miss any element in collector.
– TongChen
Jan 4 at 3:24
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
Jan 5 at 0:00
It looks like your combiner (
(a,b)->a) is missing b in the result– Marcono1234
Jan 4 at 2:52
It looks like your combiner (
(a,b)->a) is missing b in the result– Marcono1234
Jan 4 at 2:52
@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
Jan 4 at 3:24
@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
Jan 4 at 3:24
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
Jan 5 at 0:00
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
Jan 5 at 0:00
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.
answered Jan 4 at 3:36
Donald RaabDonald Raab
4,22112029
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.
answered Jan 4 at 3:36
Donald RaabDonald Raab
4,22112029
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.
answered Jan 4 at 3:36
Donald RaabDonald Raab
4,22112029
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.
answered Jan 4 at 3:36
Donald RaabDonald Raab
4,22112029
edited Jan 4 at 6:51
answered Jan 4 at 3:36
Donald RaabDonald Raab
4,22112029
answered Jan 4 at 3:36
Donald RaabDonald Raab
4,22112029
answered Jan 4 at 3:36
Donald RaabDonald Raab
4,22112029
4,22112029
add a comment |
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();
answered Jan 4 at 1:42
KeijackKeijack
1666
Thanks @Hulk, add theintversion, but I think I like theIntegerversion, simpler.
– Keijack
Jan 4 at 8:05
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();
answered Jan 4 at 1:42
KeijackKeijack
1666
Thanks @Hulk, add theintversion, but I think I like theIntegerversion, simpler.
– Keijack
Jan 4 at 8:05
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();
answered Jan 4 at 1:42
KeijackKeijack
1666
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();
answered Jan 4 at 1:42
KeijackKeijack
1666
edited Jan 4 at 8:03
answered Jan 4 at 1:42
KeijackKeijack
1666
answered Jan 4 at 1:42
KeijackKeijack
1666
answered Jan 4 at 1:42
KeijackKeijack
1666
1666
Thanks @Hulk, add theintversion, but I think I like theIntegerversion, simpler.
– Keijack
Jan 4 at 8:05
add a comment |
Thanks @Hulk, add theintversion, but I think I like theIntegerversion, simpler.
– Keijack
Jan 4 at 8:05
Thanks @Hulk, add the
int version, but I think I like the Integer version, simpler.– Keijack
Jan 4 at 8:05
Thanks @Hulk, add the
int version, but I think I like the Integer version, simpler.– Keijack
Jan 4 at 8:05
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.
edited Jan 4 at 8:01
ZhaoGang
1,9011116
answered Jan 4 at 1:53
JaiJai
5,76811231
This does not reverse the queue (or its values)
– Marcono1234
Jan 4 at 2:50
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.
edited Jan 4 at 8:01
ZhaoGang
1,9011116
answered Jan 4 at 1:53
JaiJai
5,76811231
This does not reverse the queue (or its values)
– Marcono1234
Jan 4 at 2:50
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.
edited Jan 4 at 8:01
ZhaoGang
1,9011116
answered Jan 4 at 1:53
JaiJai
5,76811231
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 Jan 4 at 8:01
ZhaoGang
1,9011116
answered Jan 4 at 1:53
JaiJai
5,76811231
edited Jan 4 at 8:01
ZhaoGang
1,9011116
edited Jan 4 at 8:01
ZhaoGang
1,9011116
edited Jan 4 at 8:01
ZhaoGang
1,9011116
1,9011116
answered Jan 4 at 1:53
JaiJai
5,76811231
answered Jan 4 at 1:53
JaiJai
5,76811231
answered Jan 4 at 1:53
JaiJai
5,76811231
5,76811231
This does not reverse the queue (or its values)
– Marcono1234
Jan 4 at 2:50
add a comment |
This does not reverse the queue (or its values)
– Marcono1234
Jan 4 at 2:50
This does not reverse the queue (or its values)
– Marcono1234
Jan 4 at 2:50
This does not reverse the queue (or its values)
– Marcono1234
Jan 4 at 2:50
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();
}
));
answered Jan 4 at 9:17
aminographyaminography
5,91321131
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();
}
));
answered Jan 4 at 9:17
aminographyaminography
5,91321131
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();
}
));
answered Jan 4 at 9:17
aminographyaminography
5,91321131
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();
}
));
answered Jan 4 at 9:17
aminographyaminography
5,91321131
edited Jan 4 at 9:33
answered Jan 4 at 9:17
aminographyaminography
5,91321131
answered Jan 4 at 9:17
aminographyaminography
5,91321131
answered Jan 4 at 9:17
aminographyaminography
5,91321131
5,91321131
add a comment |
add a comment |
Here's a way that creates a reversed array without reversing the queue:
int i = { queue.size() };
int array = new int[i[0]];
queue.forEach(n -> array[--i[0]] = n);
The above code is quite hacky due to the impossibility to modify local variables from within lambda expressions. So here i needs to be a one-element array, to overcome this restriction.
Note: bear in mind that I've come to this solution just for fun :)
answered Jan 5 at 16:53
Federico Peralta SchaffnerFederico Peralta Schaffner
22.3k43573
add a comment |
Here's a way that creates a reversed array without reversing the queue:
int i = { queue.size() };
int array = new int[i[0]];
queue.forEach(n -> array[--i[0]] = n);
The above code is quite hacky due to the impossibility to modify local variables from within lambda expressions. So here i needs to be a one-element array, to overcome this restriction.
Note: bear in mind that I've come to this solution just for fun :)
answered Jan 5 at 16:53
Federico Peralta SchaffnerFederico Peralta Schaffner
22.3k43573
add a comment |
Here's a way that creates a reversed array without reversing the queue:
int i = { queue.size() };
int array = new int[i[0]];
queue.forEach(n -> array[--i[0]] = n);
The above code is quite hacky due to the impossibility to modify local variables from within lambda expressions. So here i needs to be a one-element array, to overcome this restriction.
Note: bear in mind that I've come to this solution just for fun :)
answered Jan 5 at 16:53
Federico Peralta SchaffnerFederico Peralta Schaffner
22.3k43573
Here's a way that creates a reversed array without reversing the queue:
int i = { queue.size() };
int array = new int[i[0]];
queue.forEach(n -> array[--i[0]] = n);
The above code is quite hacky due to the impossibility to modify local variables from within lambda expressions. So here i needs to be a one-element array, to overcome this restriction.
Note: bear in mind that I've come to this solution just for fun :)
answered Jan 5 at 16:53
Federico Peralta SchaffnerFederico Peralta Schaffner
22.3k43573
edited Jan 5 at 17:06
answered Jan 5 at 16:53
Federico Peralta SchaffnerFederico Peralta Schaffner
22.3k43573
answered Jan 5 at 16:53
Federico Peralta SchaffnerFederico Peralta Schaffner
22.3k43573
answered Jan 5 at 16:53
Federico Peralta SchaffnerFederico Peralta Schaffner
22.3k43573
22.3k43573
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