java 将列表的第一个元素移到末尾

Move the first element of a list to the end

提问by elaich

is there any clever way to do that ? my best way was:

有什么聪明的方法可以做到这一点吗？我最好的方法是：

object next = list.get(0) ;
list.remove(0) ;
list.add(next) ;

if not is there any type of collection that will make that easier ? I don't like the need of a temporary object to store the element i want to move ..

如果没有，是否有任何类型的收藏可以使这更容易？我不喜欢需要一个临时对象来存储我想要移动的元素..

EDIT: I have tested the propositions listed below, with my code:

编辑：我已经用我的代码测试了下面列出的命题：

    long starttime = System.nanoTime() ;
    for (int i = 0; i < ntours; i++){
        profit += retrieveGroupsWillPlay(groups, ngroups, limit) ;
    }
    long endtime = System.nanoTime() ;
    System.out.println("Timing: " + (endtime - starttime)) ;
    System.out.println("Profit: " + profit) ;

here is the results: (profit: 15, ensure that the result is right for my code) code:

这是结果：（利润：15，确保结果适合我的代码）代码：

private static int retrieveGroupsWillPlay(ArrayList<Integer> queue,int ngroups, int limit) { 
    int peopleWillPlay = 0 ;
    for (int i = 0; i < ngroups; i++){
        int nextGroup = queue.get(0) ;
        if(limit >= peopleWillPlay + nextGroup) {
            peopleWillPlay += nextGroup ;
            queue.add(nextGroup) ;
            queue.remove(0) ;
        }
        else break ;
    }
    return peopleWillPlay ;
}

results:

结果：

Timing: 23326
Profit: 15
Timing: 22171
Profit: 15
Timing: 22156
Profit: 15
Timing: 22944
Profit: 15
Timing: 22240
Profit: 15
Timing: 21769
Profit: 15
Timing: 21866
Profit: 15
Timing: 22341
Profit: 15
Timing: 24049
Profit: 15
Timing: 22420
Profit: 15

code:

代码：

private static int retrieveGroupsWillPlay(ArrayList<Integer> queue,int ngroups, int limit) { 
    int peopleWillPlay = 0 ;
    for (int i = 0; i < ngroups; i++){
        int nextGroup = queue.get(0) ;
        if(limit >= peopleWillPlay + nextGroup) {
            peopleWillPlay += nextGroup ;
            Collections.rotate(queue, -1) ;
        }
        else break ;
    }
    return peopleWillPlay ;
}

results:

结果：

Timing: 92101
Profit: 15
Timing: 87137
Profit: 15
Timing: 84531
Profit: 15
Timing: 105919
Profit: 15
Timing: 77019
Profit: 15
Timing: 84805
Profit: 15
Timing: 93393
Profit: 15
Timing: 77079
Profit: 15
Timing: 84315
Profit: 15
Timing: 107002
Profit: 15

code:

代码：

private static int retrieveGroupsWillPlay(ArrayList<Integer> queue,int ngroups, int limit) { 
    int peopleWillPlay = 0 ;
    for (int i = 0; i < ngroups; i++){
        int nextGroup = queue.get(0) ;
        if(limit >= peopleWillPlay + nextGroup) {
            peopleWillPlay += nextGroup ;
            queue.add(queue.remove(0)) ;
        }
        else break ;
    }
    return peopleWillPlay ;
}

results:

结果：

Timing: 28079
Profit: 15
Timing: 28994
Profit: 15
Timing: 29525
Profit: 15
Timing: 22240
Profit: 15
Timing: 38326
Profit: 15
Timing: 33742
Profit: 15
Timing: 21500
Profit: 15
Timing: 22714
Profit: 15
Timing: 20939
Profit: 15
Timing: 30157
Profit: 15

code:

代码：

private static int retrieveGroupsWillPlay(LinkedList<Integer> queue,int ngroups, int limit) { 
    int peopleWillPlay = 0 ;
    for (int i = 0; i < ngroups; i++){
        int nextGroup = queue.get(0) ;
        if(limit >= peopleWillPlay + nextGroup) {
            peopleWillPlay += nextGroup ;
            queue.addLast(queue.removeFirst()) ;
        }
        else break ;
    }
    return peopleWillPlay ;
}

result:

结果：

Timing: 31104
Profit: 15
Timing: 42332
Profit: 15
Timing: 36443
Profit: 15
Timing: 31840
Profit: 15
Timing: 31387
Profit: 15
Timing: 32102
Profit: 15
Timing: 31347
Profit: 15
Timing: 30666
Profit: 15
Timing: 32781
Profit: 15
Timing: 32464
Profit: 15

code:

代码：

private static int retrieveGroupsWillPlay(LinkedList<Integer> queue,int ngroups, int limit) { 
    int peopleWillPlay = 0 ;
    for (int i = 0; i < ngroups; i++){
        int nextGroup = queue.get(0) ;
        if(limit >= peopleWillPlay + nextGroup) {
            peopleWillPlay += nextGroup ;
            queue.offer(queue.poll()) ;
        }
        else break ;
    }
    return peopleWillPlay ;
}

results:

结果：

Timing: 35389
Profit: 15
Timing: 34849
Profit: 15
Timing: 43606
Profit: 15
Timing: 41796
Profit: 15
Timing: 51122
Profit: 15
Timing: 59302
Profit: 15
Timing: 32340
Profit: 15
Timing: 35654
Profit: 15
Timing: 34586
Profit: 15
Timing: 35479
Profit: 15 

回答by dasblinkenlight

You can use Collections.rotatefor that:

你可以使用Collections.rotate：

Collections.rotate(list, -1);

回答by pcalcao

I'm not exactly sure what you want to do, but here goes:

我不确定你想做什么，但这里有：

If you're using something like an ArrayList, you can do:

如果您使用的是类似的东西ArrayList，您可以执行以下操作：

list.add(list.remove(0));

Please keep in mind that removefrom an ArrayList runs in linear time, that is, O(N), so that is extremely inefficient.

请记住，remove从 ArrayList 以线性时间运行，即O(N)，因此效率极低。

In case you can choose the type of List, you probably want a LinkedList, that implementes the Dequeueinterface, so it would allow you to do something like:

如果您可以选择 List 的类型，您可能需要一个LinkedList实现Dequeue接口的 , 这样您就可以执行以下操作：

list.offer(list.poll());

Both offerand pollare operations done in constant time.

双方offer并poll在不断的时间内完成操作。

If you want to use a builtin from the Collectionsclass, you can do as @dasblinkenlight suggested and use Collections.rotate(list, -1);(adding it here for completeness).

如果您想使用Collections类中的内置函数，您可以按照@dasblinkenlight 的建议进行操作并使用Collections.rotate(list, -1);（在此处添加以确保完整性）。

回答by codeghost

You don't need a temp variable just write:

您不需要临时变量，只需编写：

list.add(list.remove(0));

回答by Betalord

The obvious answer is:

显而易见的答案是：

list.add(list.remove(0))

list.add(list.remove(0))

which is the most elegant way, I believe. You could alternatively use

我相信这是最优雅的方式。你也可以使用

Collections.swap(list.get(0), list.get(list.size()-1))

Collections.swap(list.get(0), list.get(list.size()-1))

however that will change position of another element (the last one). You could also use Collections.rotate(list, -1), however rotating the list could mean moving all of its elements (it depends on List implementation I guess) and that might not be efficient.

但是这会改变另一个元素（最后一个）的位置。您也可以使用Collections.rotate(list, -1)，但是旋转列表可能意味着移动其所有元素（我猜这取决于 List 实现）并且这可能效率不高。

回答by PermGenError

You can also use LinkedList#addLast()method.

您还可以使用LinkedList#addLast()方法。

list.add(next) ;     
list.addLast(list.removeFirst());

回答by Rob

You want a dequeu(short for double ended queue).

你想要一个dequeu（双端队列的缩写）。