我有以下方法获取rgb值并使用较小的调色板对其进行分类:

private static int roundToNearestColor( int rgb, int nrColors )
    {
        int red = ( rgb >> 16 ) & 0xFF;
        int green = ( rgb >> 8 ) & 0xFF;
        int blue = ( rgb & 0xFF );
        red = red - ( red % nrColors );
        green = green - ( green % nrColors );
        blue = blue - ( blue % nrColors );
        return 0xFF000000 | ( red << 16 ) | ( green << 8 ) | ( blue );
    }

令我恼火的代码是
red = red - ( red % nrColors );
green = green - ( green % nrColors );
blue = blue - ( blue % nrColors );

我相信有一个替代的按位版本,它将执行得更快,但由于我的按位算法有点生疏,我有困难找到这样的表达式。任何帮助或意见将不胜感激。

最佳答案

如果nrColors总是2的幂:

private static int roundToNearestColor( int rgb, int nrColors )
{
    if (Integer.bitCount(nrColors) != 1) {
        throw new IllegalArgumentException("nrColors must be a power of two");
    }
    int mask = 0xFF & (-1 << Integer.numberOfTrailingZeros(nrColors));
    int red = ( rgb >> 16 ) & mask;
    int green = ( rgb >> 8 ) & mask;
    int blue = ( rgb & mask );
    return 0xFF000000 | ( red << 16 ) | ( green << 8 ) | ( blue );
}

关于java - 在Java中查找RGB的按位版本,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/4783529/

10-09 08:49