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                合規國際互聯網加速 OSASE為企業客戶提供高速穩定SD-WAN國際加速解決方案。 廣告
                關于銳化效果,感覺效果并沒有達到預期的效果。 處理后感覺像素變的粗糙了。 ![](https://box.kancloud.cn/2016-03-28_56f8ea7abfc72.jpg) 這涉及到數學知識,是關于拉普拉斯矩陣的應用,而本人在數學方面并不牛,所以不敢在這細講,怕誤導大家。 ~~~ long start = System.currentTimeMillis(); // 拉普拉斯矩陣 int[] laplacian = new int[] { -1, -1, -1, -1, 9, -1, -1, -1, -1 }; int width = bmp.getWidth(); int height = bmp.getHeight(); Bitmap bitmap = Bitmap.createBitmap(width, height, Bitmap.Config.RGB_565); int pixR = 0; int pixG = 0; int pixB = 0; int pixColor = 0; int newR = 0; int newG = 0; int newB = 0; int idx = 0; float alpha = 0.3F; int[] pixels = new int[width * height]; bmp.getPixels(pixels, 0, width, 0, 0, width, height); ~~~ //同樣,和上一篇一樣,對像素點的RGB做處理 不過這里要用上拉普拉斯矩陣,做信息修改。 以下是對像素的詳細操作核心: ~~~ for (int i = 1, length = height - 1; i < length; i++) { for (int k = 1, len = width - 1; k < len; k++) { idx = 0; for (int m = -1; m <= 1; m++) { for (int n = -1; n <= 1; n++) { pixColor = pixels[(i + n) * width + k + m]; pixR = Color.red(pixColor); pixG = Color.green(pixColor); pixB = Color.blue(pixColor); newR = newR + (int) (pixR * laplacian[idx] * alpha); newG = newG + (int) (pixG * laplacian[idx] * alpha); newB = newB + (int) (pixB * laplacian[idx] * alpha); idx++; } } newR = Math.min(255, Math.max(0, newR)); newG = Math.min(255, Math.max(0, newG)); newB = Math.min(255, Math.max(0, newB)); pixels[i * width + k] = Color.argb(255, newR, newG, newB); newR = 0; newG = 0; newB = 0; } } //最后將結果返回 bitmap.setPixels(pixels, 0, width, 0, 0, width, height); long end = System.currentTimeMillis(); Log.d("may", "used time="+(end - start)); return bitmap; ~~~ 這樣,銳化效果就達到了,雖然效果并不明顯,本人還在優化當中。
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