Reversing a Queue and converting it into an int array












8














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:




  1. the explict casting (List)queue;

  2. 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.










share|improve this question





























    8














    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:




    1. the explict casting (List)queue;

    2. 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.










    share|improve this question



























      8












      8








      8


      3





      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:




      1. the explict casting (List)queue;

      2. 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.










      share|improve this question















      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:




      1. the explict casting (List)queue;

      2. 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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited yesterday









      Moira

      5,25221937




      5,25221937










      asked yesterday









      ZhaoGang

      1,6181015




      1,6181015
























          9 Answers
          9






          active

          oldest

          votes


















          4














          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());





          share|improve this answer





















          • but this wouldn't reverse the queue.
            – nullpointer
            yesterday








          • 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



















          6














          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.






          share|improve this answer





























            5














            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]);
            }





            share|improve this answer























            • 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






            • 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
              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
              yesterday



















            3














            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();





            share|improve this answer





















            • 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










            • 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
              22 hours ago



















            2














            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();





            share|improve this answer























            • Thanks @Hulk, add the int version, but I think I like the Integer version, simpler.
              – Keijack
              yesterday



















            1














            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.






            share|improve this answer































              1














              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.






              share|improve this answer























              • This does not reverse the queue (or its values)
                – Marcono1234
                yesterday



















              0














              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();
              }
              ));





              share|improve this answer































                0














                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 :)






                share|improve this answer























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                  9 Answers
                  9






                  active

                  oldest

                  votes








                  9 Answers
                  9






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  4














                  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());





                  share|improve this answer





















                  • but this wouldn't reverse the queue.
                    – nullpointer
                    yesterday








                  • 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
















                  4














                  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());





                  share|improve this answer





















                  • but this wouldn't reverse the queue.
                    – nullpointer
                    yesterday








                  • 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














                  4












                  4








                  4






                  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());





                  share|improve this answer












                  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());






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered yesterday









                  Elliott Frisch

                  153k1389178




                  153k1389178












                  • but this wouldn't reverse the queue.
                    – nullpointer
                    yesterday








                  • 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


















                  • but this wouldn't reverse the queue.
                    – nullpointer
                    yesterday








                  • 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
















                  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













                  6














                  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.






                  share|improve this answer


























                    6














                    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.






                    share|improve this answer
























                      6












                      6








                      6






                      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.






                      share|improve this answer












                      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.







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered yesterday









                      xehpuk

                      4,3072335




                      4,3072335























                          5














                          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]);
                          }





                          share|improve this answer























                          • 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






                          • 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
                            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
                            yesterday
















                          5














                          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]);
                          }





                          share|improve this answer























                          • 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






                          • 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
                            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
                            yesterday














                          5












                          5








                          5






                          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]);
                          }





                          share|improve this answer














                          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]);
                          }






                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited yesterday

























                          answered yesterday









                          nullpointer

                          43.3k1093178




                          43.3k1093178












                          • 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






                          • 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
                            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
                            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






                          • 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
                            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
                            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




                          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




                          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
                          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
                          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
                          yesterday











                          3














                          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();





                          share|improve this answer





















                          • 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










                          • 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
                            22 hours ago
















                          3














                          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();





                          share|improve this answer





















                          • 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










                          • 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
                            22 hours ago














                          3












                          3








                          3






                          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();





                          share|improve this answer












                          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();






                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered yesterday









                          TongChen

                          1958




                          1958












                          • 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










                          • 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
                            22 hours ago


















                          • 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










                          • 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
                            22 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
                          22 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
                          22 hours ago











                          2














                          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();





                          share|improve this answer























                          • Thanks @Hulk, add the int version, but I think I like the Integer version, simpler.
                            – Keijack
                            yesterday
















                          2














                          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();





                          share|improve this answer























                          • Thanks @Hulk, add the int version, but I think I like the Integer version, simpler.
                            – Keijack
                            yesterday














                          2












                          2








                          2






                          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();





                          share|improve this answer














                          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();






                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited yesterday

























                          answered yesterday









                          Keijack

                          1666




                          1666












                          • Thanks @Hulk, add the int version, but I think I like the Integer version, simpler.
                            – Keijack
                            yesterday


















                          • Thanks @Hulk, add the int version, but I think I like the Integer version, simpler.
                            – Keijack
                            yesterday
















                          Thanks @Hulk, add the int version, but I think I like the Integer version, simpler.
                          – Keijack
                          yesterday




                          Thanks @Hulk, add the int version, but I think I like the Integer version, simpler.
                          – Keijack
                          yesterday











                          1














                          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.






                          share|improve this answer




























                            1














                            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.






                            share|improve this answer


























                              1












                              1








                              1






                              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.






                              share|improve this answer














                              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.







                              share|improve this answer














                              share|improve this answer



                              share|improve this answer








                              edited yesterday

























                              answered yesterday









                              Donald Raab

                              4,21112029




                              4,21112029























                                  1














                                  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.






                                  share|improve this answer























                                  • This does not reverse the queue (or its values)
                                    – Marcono1234
                                    yesterday
















                                  1














                                  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.






                                  share|improve this answer























                                  • This does not reverse the queue (or its values)
                                    – Marcono1234
                                    yesterday














                                  1












                                  1








                                  1






                                  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.






                                  share|improve this answer














                                  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.







                                  share|improve this answer














                                  share|improve this answer



                                  share|improve this answer








                                  edited yesterday









                                  ZhaoGang

                                  1,6181015




                                  1,6181015










                                  answered yesterday









                                  Jai

                                  5,73311231




                                  5,73311231












                                  • 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




                                  This does not reverse the queue (or its values)
                                  – Marcono1234
                                  yesterday











                                  0














                                  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();
                                  }
                                  ));





                                  share|improve this answer




























                                    0














                                    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();
                                    }
                                    ));





                                    share|improve this answer


























                                      0












                                      0








                                      0






                                      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();
                                      }
                                      ));





                                      share|improve this answer














                                      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();
                                      }
                                      ));






                                      share|improve this answer














                                      share|improve this answer



                                      share|improve this answer








                                      edited yesterday

























                                      answered yesterday









                                      aminography

                                      5,51821130




                                      5,51821130























                                          0














                                          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 :)






                                          share|improve this answer




























                                            0














                                            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 :)






                                            share|improve this answer


























                                              0












                                              0








                                              0






                                              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 :)






                                              share|improve this answer














                                              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 :)







                                              share|improve this answer














                                              share|improve this answer



                                              share|improve this answer








                                              edited 5 hours ago

























                                              answered 5 hours ago









                                              Federico Peralta Schaffner

                                              22k43370




                                              22k43370






























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