java 从整数的可迭代对象中获取排序数组的最快方法
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Fastest way to get a sorted array out of an Iterable of integers
提问by Rajat Gupta
Possible Duplicate:
Fastest strategy to form and sort an array of positive integers
可能的重复：
形成和排序正整数数组的最快策略
What would be the fastest way to get a sorted array from an unsorted iterable of integers ? Currently I do it by iterating over the iterable n no of times (where n is size of list) each time getting the highest from iterable & putting it in array. But I'm looking to clean this up & let some good library do it for me.
从未排序的整数迭代中获取排序数组的最快方法是什么？目前，我通过迭代可迭代 n 次（其中 n 是列表的大小）来实现，每次从可迭代中获取最高值并将其放入数组中。但我正在寻找清理它并让一些好的图书馆为我做这件事。
Probably I won't mind using any popular libraries like Guava, etc for this purpose.
可能我不介意为此目的使用任何流行的库，例如 Guava 等。
采纳答案by assylias
As commented, the easiest and probably fastest way is to populate a collection and sort it - I would simply use an ArrayList in this case:
正如所评论的，最简单也可能是最快的方法是填充一个集合并对其进行排序 - 在这种情况下，我将简单地使用 ArrayList：
List<Integer> sortedList = new ArrayList<>();
for (Integer i : yourIterable) {
sortedList.add(i);
}
Collections.sort(sortedList);
If you know the size of your Iterable beforehand you can initialise the arraylist with the right size to make an additional efficiency gain:
如果您事先知道 Iterable 的大小，您可以使用正确的大小初始化 arraylist 以提高效率：
List<Integer> sortedList = new ArrayList<>(size);
回答by Henrik Aasted S?rensen
It is essentially the same answer as assyliashas already provided, but if you have Google Guava on the classpath, you can shorten it to :
它与assylias已经提供的答案基本相同，但是如果类路径上有 Google Guava，则可以将其缩短为：
import java.util.Collections;
import java.util.List;
import com.google.common.collect.Lists;
...
List list = Lists.newArrayList(iterable);
Collections.sort(list);
