尝试对图像进行广度优先搜索，（或者遍历点列表），并标记与另一个像素相邻的每个像素不属于同一组。
我不清楚你想要的周长是请求对象外边缘上的每个像素，还是对象边界的每个像素。我暂时假设前者。
设置图片：
这就是你如何去做的。首先，将图像设置为二维数组，每个像素都标有组号：
[ 0 0 0 0 1 1 0 ]
[ 0 0 2 0 0 0 0 ]
[ 0 2 2 0 3 3 0 ]
[ 0 2 2 0 0 3 0 ]
[ 0 2 2 0 0 0 0 ]
[ 0 2 0 0 4 4 0 ]
[ 0 0 0 0 4 4 0 ]
加载它的一个好方法是使用Scanner 对象逐个获取每个点：
List<Point> points = new ArrayList<>();
Scanner scanner = new Scanner( /* whatever your input source is */ );
String pointRegex = "\\(\\d,\\d\\)"; //looks for something like "(#,#)"
while(!scanner.hasNext(pointRegex)){
String pointText = scanner.next(pointRegex); //For example, "(5,4)"
Point point = getPointFromText(pointText); //turns a string into a point
points.add(point);
}
注意Scanner.next(String pattern) 的使用。这个方法将返回下一个看起来像该模式的String。 （如果您想了解更多有关其工作原理的信息，请阅读正则表达式。）
现在开始填充网格：
boolean[][] binaryImage = new boolean[width][height];
for(Point p : points){ //Iterate through each Point inside our List of Point objects
binaryImage[p.getX()][p.getY()] = true;
}
这会将由我们的Point 对象集合“points”表示的对象放入booleans 的网格中。我们只需要担心这个对象，所以我们不需要加载任何其他对象。现在要找出周长上的点。
递归方法：
boolean[][] visitedBefore = new boolean[width][height];
boolean[][] isOnPerimeter = new boolean[width][height];
int[] deltaX = {-1, 0, 1, -1, 1, -1, 0, 1},
deltaY = {-1, -1, -1, 0, 0, 1, 1, 1};
Queue<Point> searchNext = new LinkedList<>();
searchNext.add(points.get(0)); //Just need one point to get going
while(!searchNext.isEmpty()){
Point p = searchNext.remove(); //take what's waiting at the front of the queue
if(visitedBefore[p.getX()][p.getY()]){
continue; //already check this spot!
}
//mark that we've been here
visited[p.getX()][p.getY()] = true;
//look at all of this Point's neighbors
for(int i = 0 ; i < deltaX.length ; i++){
int newX = p.getX() + deltaX[i];
int newY = p.getY() + deltaY[i];
//make sure this isn't out of bounds
if(newX < 0 || newX >= width || newY<0 || newY>=height){
isOnPerimeter[p.getX()][p.getY()] = true; //if you decide bordering the edge of the image counts as being on the perimeter
continue;
}
//check if this new point we're considering isn't part of the image
if( binaryImage[p.getX()][p.getY()] != binaryImage[newX][newY] ){
//if it isn't, then this Point p must be on the perimeter
isOnPerimeter[p.getX()][p.getY()] = true;
} else {
/* otherwise, this new point we're considering is part of the
* same object, and could be part of the perimeter. */
searchNext.add(new Point(newX, newY));
}
}
}
现在您有一个网格，周长上的每个点都标记为true。如果您需要这些作为列表，那么挑选这些要点很容易：
List<Point> perimeter = new ArrayList<Point>();
for(int x = 0 ; x < isOnPerimeter.length ; x++)
for(int y = 0 ; y < isOnPerimeter[x].length ; y++)
perimeter.add( new Point(x,y) );
迭代法：
这与上面的非常相似，但直接将周长点放入列表中。
int[] deltaX = {-1, 0, 1, -1, 1, -1, 0, 1},
deltaY = {-1, -1, -1, 0, 0, 1, 1, 1};
outer: for(Point p : points){
inner: for(int i = 0 ; i < deltaX.length ; i++){
int newX = p.getX() + deltaX[i];
int newY = p.getY() + deltaY[i];
//check if this new point we're considering is outside the image
if(newX < 0 || newX >= width || newY<0 || newY>=height){
perimeter.add(p); //if you decide bordering the edge of the image counts as being on the perimeter
continue outer;
}
//check if this new point we're considering isn't part of the image
if( binaryImage[p.getX()][p.getY()] != binaryImage[newX][newY] ){
//if it isn't, then this Point p must be on the perimeter
perimeter.add(p);
continue outer;
}
}
}
注意标签outer: 和inner:。当我们说continue outer; 时，这让我们可以选择跳过哪个 for 循环。
给你！这应该可以帮助您以二进制图像或列表的形式获取任何对象的周长。