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Android毛玻璃处理代码(Blur)

2016-03-14 11:30 471 查看
  以下为将bitmap图像处理为毛玻璃效果的图像的工具类:

public class FastBlurUtil {

public static Bitmap doBlur(Bitmap sentBitmap, int radius, boolean canReuseInBitmap) {

// Stack Blur v1.0 from
// http://www.quasimondo.com/StackBlurForCanvas/StackBlurDemo.html //
// Java Author: Mario Klingemann <mario at quasimondo.com>
// http://incubator.quasimondo.com // created Feburary 29, 2004
// Android port : Yahel Bouaziz <yahel at kayenko.com>
// http://www.kayenko.com // ported april 5th, 2012

// This is a compromise between Gaussian Blur and Box blur
// It creates much better looking blurs than Box Blur, but is
// 7x faster than my Gaussian Blur implementation.
//
// I called it Stack Blur because this describes best how this
// filter works internally: it creates a kind of moving stack
// of colors whilst scanning through the image. Thereby it
// just has to add one new block of color to the right side
// of the stack and remove the leftmost color. The remaining
// colors on the topmost layer of the stack are either added on
// or reduced by one, depending on if they are on the right or
// on the left side of the stack.
//
// If you are using this algorithm in your code please add
// the following line:
//
// Stack Blur Algorithm by Mario Klingemann <mario@quasimondo.com>

Bitmap bitmap;
if (canReuseInBitmap) {
bitmap = sentBitmap;
} else {
bitmap = sentBitmap.copy(sentBitmap.getConfig(), true);
}

if (radius < 1) {
return (null);
}

int w = bitmap.getWidth();
int h = bitmap.getHeight();

int[] pix = new int[w * h];
bitmap.getPixels(pix, 0, w, 0, 0, w, h);

int wm = w - 1;
int hm = h - 1;
int wh = w * h;
int div = radius + radius + 1;

int r[] = new int[wh];
int g[] = new int[wh];
int b[] = new int[wh];
int rsum, gsum, bsum, x, y, i, p, yp, yi, yw;
int vmin[] = new int[Math.max(w, h)];

int divsum = (div + 1) >> 1;
divsum *= divsum;
int dv[] = new int[256 * divsum];
for (i = 0; i < 256 * divsum; i++) {
dv[i] = (i / divsum);
}

yw = yi = 0;

int[][] stack = new int[div][3];
int stackpointer;
int stackstart;
int[] sir;
int rbs;
int r1 = radius + 1;
int routsum, goutsum, boutsum;
int rinsum, ginsum, binsum;

for (y = 0; y < h; y++) {
rinsum = ginsum = binsum = routsum = goutsum = boutsum = rsum = gsum = bsum = 0;
for (i = -radius; i <= radius; i++) {
p = pix[yi + Math.min(wm, Math.max(i, 0))];
sir = stack[i + radius];
sir[0] = (p & 0xff0000) >> 16;
sir[1] = (p & 0x00ff00) >> 8;
sir[2] = (p & 0x0000ff);
rbs = r1 - Math.abs(i);
rsum += sir[0] * rbs;
gsum += sir[1] * rbs;
bsum += sir[2] * rbs;
if (i > 0) {
rinsum += sir[0];
ginsum += sir[1];
binsum += sir[2];
} else {
routsum += sir[0];
goutsum += sir[1];
boutsum += sir[2];
}
}
stackpointer = radius;

for (x = 0; x < w; x++) {

r[yi] = dv[rsum];
g[yi] = dv[gsum];
b[yi] = dv[bsum];

rsum -= routsum;
gsum -= goutsum;
bsum -= boutsum;

stackstart = stackpointer - radius + div;
sir = stack[stackstart % div];

routsum -= sir[0];
goutsum -= sir[1];
boutsum -= sir[2];

if (y == 0) {
vmin[x] = Math.min(x + radius + 1, wm);
}
p = pix[yw + vmin[x]];

sir[0] = (p & 0xff0000) >> 16;
sir[1] = (p & 0x00ff00) >> 8;
sir[2] = (p & 0x0000ff);

rinsum += sir[0];
ginsum += sir[1];
binsum += sir[2];

rsum += rinsum;
gsum += ginsum;
bsum += binsum;

stackpointer = (stackpointer + 1) % div;
sir = stack[(stackpointer) % div];

routsum += sir[0];
goutsum += sir[1];
boutsum += sir[2];

rinsum -= sir[0];
ginsum -= sir[1];
binsum -= sir[2];

yi++;
}
yw += w;
}
for (x = 0; x < w; x++) {
rinsum = ginsum = binsum = routsum = goutsum = boutsum = rsum = gsum = bsum = 0;
yp = -radius * w;
for (i = -radius; i <= radius; i++) {
yi = Math.max(0, yp) + x;

sir = stack[i + radius];

sir[0] = r[yi];
sir[1] = g[yi];
sir[2] = b[yi];

rbs = r1 - Math.abs(i);

rsum += r[yi] * rbs;
gsum += g[yi] * rbs;
bsum += b[yi] * rbs;

if (i > 0) {
rinsum += sir[0];
ginsum += sir[1];
binsum += sir[2];
} else {
routsum += sir[0];
goutsum += sir[1];
boutsum += sir[2];
}

if (i < hm) {
yp += w;
}
}
yi = x;
stackpointer = radius;
for (y = 0; y < h; y++) {
// Preserve alpha channel: ( 0xff000000 & pix[yi] )
pix[yi] = (0xff000000 & pix[yi]) | (dv[rsum] << 16) | (dv[gsum] << 8) | dv[bsum];

rsum -= routsum;
gsum -= goutsum;
bsum -= boutsum;

stackstart = stackpointer - radius + div;
sir = stack[stackstart % div];

routsum -= sir[0];
goutsum -= sir[1];
boutsum -= sir[2];

if (x == 0) {
vmin[y] = Math.min(y + r1, hm) * w;
}
p = x + vmin[y];

sir[0] = r[p];
sir[1] = g[p];
sir[2] = b[p];

rinsum += sir[0];
ginsum += sir[1];
binsum += sir[2];

rsum += rinsum;
gsum += ginsum;
bsum += binsum;

stackpointer = (stackpointer + 1) % div;
sir = stack[stackpointer];

routsum += sir[0];
goutsum += sir[1];
boutsum += sir[2];

rinsum -= sir[0];
ginsum -= sir[1];
binsum -= sir[2];

yi += w;
}
}

bitmap.setPixels(pix, 0, w, 0, 0, w, h);

return (bitmap);
}

}


  可以看出,使用方法非常简单,传入待续话的bitmap、虚化程序(一般为8),和是否重用flag。

  然后,对于大图,往往会出现OOM异常的报错,那是因为当虚化开始时,虚拟机开始不断进行内存回收,包括所有软引用的回收。然后,仍然出现了内存溢出。那就意味着我们只能对小图进行虚化,这样才能防止内存溢出。但我并不想换其他图,那么,我们应该把这张图先进性缩放。

缩放方法:

public static Bitmap createScaledBitmap(Bitmap src, int dstWidth, int dstHeight, boolean filter)


  第四个输入输入的参数filter,是指缩放边缘效果,filter为true则会得到一个边缘平滑的bitmap,反之,则会得到一个边缘锯齿、pixelrelated的bitmap。这里,我们对图片进行虚化,无所谓边缘效果,所以filter=false。

  那么,怎么组合运行虚化代码实现大图虚化呢?以下为具体实现代码:

int scaleRatio = 10;
int blurRadius = 8;
Bitmap scaledBitmap = Bitmap.createScaledBitmap(originBitmap,
originBitmap.getWidth() / scaleRatio,
originBitmap.getHeight() / scaleRatio,
false);
Bitmap blurBitmap = FastBlur.doBlur(scaledBitmap, blurRadius, true);
imageView.setScaleType(ImageView.ScaleType.CENTER_CROP);
imageView.setImageBitmap(blurBitmap);


  如果图片的虚化效果较弱或者并不是很明显,提供增强虚化效果的方法:

1、增大scaleRatio的数值(增大缩放比),使用更小的bitmap去虚化,这样可以得到更好的虚化效果,而且有利于减小内存消耗。

2、增大blurRadius的数值,这样可以提高续话层度,不过会导致cpu更加intensive。

参考资料:实现虚化bitmap代码地址
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