java - 有没有一种快速的方法可以将同一个数组以不同的偏移量复制到另一个数千次？

Question

我正在尝试创建声音混响效果。它工作得很好，但有一个问题：它以不同的偏移量和不同的音量复制一个巨大的数组数千次，并且需要 10 分钟来处理一个 4 分钟音频样本的小混响效果。有没有办法加快这个过程？如果需要，我可以轻松列出所有延迟。

编辑：这是我的代码。有没有更好的方法来达到相同或相似的结果？SoundSource 是一个包含声音和位置的类。WaveNode 包含一个位置以及 WavePath 实例的 ArrayList，这些实例定义了有关回波的各种参数。请告诉我是否还有其他不清楚的地方。

public static double[][] render(int sampleLength, int sampleRate, int depth) //depth is number of echoes, must be at least 0
{
    init(sampleLength, sampleRate);

    for (int iSource = 0; iSource < sourceList.size(); iSource++)
    {
        renderNode(sourceList.get(iSource).getSoundSource(), sourceList.get(iSource), 0, depth);
    }

    double[][] r = sampleData.clone();
    sampleData = null;
    return r;
}

private static void renderNode(SoundSource source, WaveNode node, int curDepth, int maxDepth)
{
    SoundSource newSource;
    WaveNode newNode;
    WavePath path;

    addSource(source.clone(), node.smoothing, node.decay);

    if (curDepth < maxDepth)
    {
        for (int i = 0; i < node.getPathList().size(); i++)
        {
            path = node.getPathList().get(i);
            newNode = findReflection(path.destId);
            newSource = SoundSource.reflection(SoundSource.swapData(source, Filter.amplify(source.getSampleData(), path.decay)), newNode.getPos(), path.smoothing);
            renderNode(newSource, newNode, curDepth + 1, maxDepth);
        }
    }
}

private static void addSource(SoundSource source, double smoothing, double decay)
{
    Vector3d pos, pos2;
    double dist, volume, soundPos, sample;
    int delay;
    double[] sourceData;

    for (int channel = 0; channel < 2; channel++)
    {
        pos = source.getPos();
        pos2 = new Vector3d(pos.x, pos.y, pos.z);
        pos2.x += earDist / 2 - earDist * channel;

        dist = Math.max(pos2.length() + source.getSourceDist(), 0.000001);
        volume = decay / dist;
        delay = (int)(dist / 343.2 * sampleRate);

        sourceData = source.getSampleData();

        sample = 0;

        for (int j = 0; j < Math.min(sourceData.length, sampleLength - delay); j++)
        {
            sample = sourceData[j] / smoothing + sample * (1 - 1 / smoothing);
            sampleData[channel][j + delay] += sample * volume;
        }
    }
}