
有一个简单的解决方案——扫描每个像素。下面的算法具有恒定的性能 O(w•h)
。
private static BufferedImage trimImage(BufferedImage image) {
int width = image.getWidth();
int height = image.getHeight();
int top = height / 2;
int bottom = top;
int left = width / 2 ;
int right = left;
for (int x = 0; x < width; x++) {
for (int y = 0; y < height; y++) {
if (image.getRGB(x, y) != 0){
top = Math.min(top, y);
bottom = Math.max(bottom, y);
left = Math.min(left, x);
right = Math.max(right, x);
}
}
}
return image.getSubimage(left, top, right - left + 1, bottom - top + 1);
}
但这更有效:
private static BufferedImage trimImage(BufferedImage image) {
WritableRaster raster = image.getAlphaRaster();
int width = raster.getWidth();
int height = raster.getHeight();
int left = 0;
int top = 0;
int right = width - 1;
int bottom = height - 1;
int minRight = width - 1;
int minBottom = height - 1;
top:
for (;top < bottom; top++){
for (int x = 0; x < width; x++){
if (raster.getSample(x, top, 0) != 0){
minRight = x;
minBottom = top;
break top;
}
}
}
left:
for (;left < minRight; left++){
for (int y = height - 1; y > top; y--){
if (raster.getSample(left, y, 0) != 0){
minBottom = y;
break left;
}
}
}
bottom:
for (;bottom > minBottom; bottom--){
for (int x = width - 1; x >= left; x--){
if (raster.getSample(x, bottom, 0) != 0){
minRight = x;
break bottom;
}
}
}
right:
for (;right > minRight; right--){
for (int y = bottom; y >= top; y--){
if (raster.getSample(right, y, 0) != 0){
break right;
}
}
}
return image.getSubimage(left, top, right - left + 1, bottom - top + 1);
}
该算法遵循pepan的答案(见上文)的想法,并且效率提高了 2 到 4 倍。不同之处在于:它从不扫描任何像素两次,并尝试在每个阶段缩小搜索范围。
方法在最坏情况下的性能是O(w•h–a•b)