[英]Move the first element of a list to the end
is there any clever way to do that ? 有什么聪明的方法吗? my best way was: 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 ? 如果没有,有没有什么类型的集合可以使这一切变得容易? I don't like the need of a temporary object to store the element i want to move .. 我不喜欢需要临时对象来存储要移动的元素。
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: 结果如下:(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: 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: 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: 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: 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
您可以为此使用Collections.rotate
:
Collections.rotate(list, -1);
I'm not exactly sure what you want to do, but here goes: 我不确定您要做什么,但是可以这样:
If you're using something like an ArrayList
, you can do: 如果您使用ArrayList
类的东西,则可以执行以下操作:
list.add(list.remove(0));
Please keep in mind that remove
from an ArrayList runs in linear time, that is, O(N)
, so that is extremely inefficient. 请记住,从ArrayList中remove
是线性时间(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 Dequeue
interface, so it would allow you to do something like: 如果您可以选择List的类型,则可能需要一个LinkedList
,它实现了Dequeue
接口,因此它将允许您执行以下操作:
list.offer(list.poll());
Both offer
and poll
are operations done in constant time. offer
和poll
都是在固定时间内完成的操作。
If you want to use a builtin from the Collections
class, you can do as @dasblinkenlight suggested and use Collections.rotate(list, -1);
如果要使用Collections
类中的内置函数,可以按照@dasblinkenlight的建议进行操作,并使用Collections.rotate(list, -1);
;。 (adding it here for completeness). (为了完整性起见,请在此处添加)。
您不需要临时变量,只需编写:
list.add(list.remove(0));
The obvious answer is: 显而易见的答案是:
list.add(list.remove(0))
which is the most elegant way, I believe. 我相信这是最优雅的方式。 You could alternatively use 您也可以使用
Collections.swap(list.get(0), list.get(list.size()-1))
however that will change position of another element (the last one). 但是,这将改变另一个元素(最后一个)的位置。 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. 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(这取决于我猜想的List实现),但这可能并不高效。
您需要出队 (双端队列的缩写)。
You can also use LinkedList#addLast() method. 您也可以使用LinkedList#addLast()方法。
list.add(next) ;
list.addLast(list.removeFirst());
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