我有一个大原始类型数组(double).如何按降序对元素进行排序?
I've got a large array of primitive types (double). How do I sort the elements in descending order?
遗憾的是,Java API 不支持使用 Comparator 对原始类型进行排序.
Unfortunately the Java API doesn't support sorting of primitive types with a Comparator.
可能想到的第一种方法是将其转换为对象列表(装箱):
The first approach that probably comes to mind is to convert it to a list of objects (boxing):
double[] array = new double[1048576]; Arrays.stream(array).boxed().sorted(Collections.reverseOrder())…但是,对数组中的每个原语进行装箱速度太慢,导致很大的 GC 压力!
However, boxing each primitive in the array is too slow and causes a lot of GC pressure!
另一种方法是排序然后反转:
Another approach would be to sort and then reverse:
double[] array = new double[1048576]; ... Arrays.sort(array); // reverse the array for (int i = 0; i < array.length / 2; i++) { // swap the elements double temp = array[i]; array[i] = array[array.length - (i + 1)]; array[array.length - (i + 1)] = temp; }这种方法也很慢 - 特别是如果数组已经排序得很好.
This approach is also slow - particularly if the array is already sorted quite well.
什么是更好的选择?
推荐答案Java Primitive 包括基于自定义比较器对原始数组进行排序的功能.使用它和 Java 8,您的示例可以编写为:
Java Primitive includes functionality for sorting primitive arrays based on a custom comparator. Using it, and Java 8, your sample could be written as:
double[] array = new double[1048576]; ... Primitive.sort(array, (d1, d2) -> Doublepare(d2, d1), false);如果您使用的是 Maven,则可以将其包含在:
If you're using Maven, you can include it with:
<dependency> <groupId>net.mintern</groupId> <artifactId>primitive</artifactId> <version>1.2.1</version> </dependency>当您将 false 作为第三个参数传递给 sort 时,它使用不稳定的排序,Java 内置 双轴快速排序.这意味着速度应该接近内置排序的速度.
When you pass false as the third argument to sort, it uses an unstable sort, a simple edit of Java's built-in dual-pivot quicksort. This means that the speed should be close to that of built-in sorting.
完全披露:我编写了 Java Primitive 库.
Full disclosure: I wrote the Java Primitive library.
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按降序对基本类型数组进行排序
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