什么是重新排序数组的最快方法 [英] What is the Fastest way to reorder an array

问题描述

什么是最快的（实时数据处理应用程序）的方式来重新排序（新的指数都是一样的），Java中的数组：

What is the fastest (realtime data processing application) way to reorder (the new indices are always the same) an array in JAVA:

例如为：
我有：
    双击[] A =新的双[] {1，234，12,99,0};

我需要得到迅速：
    双击[] B =新的双[] {A [2]，A [4]，A [0]，A [1]，A [3]};

但是，也许这是这是最有效的方式做到这一点呢？

But maybe this is this is the most efficient way to do it anyway?

非常感谢您的反馈意见

推荐答案

我怀疑你能比你目前的做法做的更好。

I doubt you can do better than your current approach of

double[] B = new double[] {A[2], A[4], A[0], A[1], A[3]};

对于其他序列可能的候选人可能是 Arrays.copyOf 或 Arrays.copyOfRange 的形式，但最低金额工作必须在这里做的包括：

Likely candidates for other sequences might be forms of Arrays.copyOf or Arrays.copyOfRange, but the minimum amount of work you must do here consists of:

• 创建一个新的数组


• 随机访问每个元件阵列中

有一个小的机会，你可能会很具体的阅读稍微好一点的做/写命令（以充分利用高速缓存行），一个猜测是什么，才能完全读取并以升序写道几乎全部：

There's a small chance that you might do slightly better with very specific read/write orders (to take advantage of cache lines), one guess is something that reads entirely in order and writes almost entire in ascending order:

   double[] B = new double[A.length];
   B[2] = A[0];
   B[3] = A[1];
   B[4] = A[3];
   B[0] = A[2];
   B[1] = A[4];

但我没有这个是明显好强的预期。如果你在那里你试图消除或优化的L1 / L2缓存命中点，是时候开始微标，而真正的答案是你应该尝试。

But I don't have strong expectations of this being noticeably better. If you're at the point where you're trying to eliminate or optimize on L1/L2 cache hits, it's time to start micro-benchmarking, and the real answer is you should experiment.

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