目前正在与Graham’s Scan和凸面HUll一起进行。我是一名学生，所以我正在努力自己完成这项工作，但我一直在筛选多个网站以找到答案。简而言之，我有我的构造函数，一个来自文件，另一个是随机生成的，这样我就可以创建一个点数组。下一步是实现快速排序，即按极轴角度进行排序。这是通过比较器类完成的。比较器类是我陷入困境的地方，我们被告知使用点比较和交叉比较来进行角度比较，但我很困惑。
/**
* Use cross product and dot product to implement this method. Do not take square roots
* or use trigonometric functions. See the PowerPoint notes on how to carry out cross and
* dot products.
*
* Call comparePolarAngle() and compareDistance().
*
* @param p1
* @param p2
* @return -1 if one of the following three conditions holds:
* a) p1 and referencePoint are the same point but p2 is a different point;
* b) neither p1 nor p2 equals referencePoint, and the polar angle of
* p1 with respect to referencePoint is less than that of p2;
* c) neither p1 nor p2 equals referencePoint, p1 and p2 have the same polar
* angle w.r.t. referencePoint, and p1 is closer to referencePoint than p2.
* 0 if p1 and p2 are the same point
* 1 if one of the following three conditions holds:
* a) p2 and referencePoint are the same point but p1 is a different point;
* b) neither p1 nor p2 equals referencePoint, and the polar angle of
* p1 with respect to referencePoint is greater than that of p2;
* c) neither p1 nor p2 equals referencePoint, p1 and p2 have the same polar
* angle w.r.t. referencePoint, and p1 is further to referencePoint than p2.
*
*/
public int compare(Point p1, Point p2){
if(p1 == referencePoint && p2 != referencePoint){
return -1;
} else if(p1 == p2){
return 0;
} else {
}
return 0;
}
/**
* Compare the polar angles of two points p1 and p2 with respect to referencePoint. Use
* cross products. Do not use trigonometric functions.
*
* Precondition: p1 and p2 are distinct points.
*
* @param p1
* @param p2
* @return -1 if p1 equals referencePoint or its polar angle with respect to referencePoint
* is less than that of p2.
* 0 if p1 and p2 have the same polar angle.
* 1 if p2 equals referencePoint or its polar angle with respect to referencePoint
* is less than that of p1.
*/
public int comparePolarAngle(Point p1, Point p2){
// TODO
return 0;
}
/**
* Compare the distances of two points p1 and p2 to referencePoint. Use dot products.
* Do not take square roots.
*
* @param p1
* @param p2
* @return -1 if p1 is closer to referencePoint
* 0 if p1 and p2 are equidistant to referencePoint
* 1 if p2 is closer to referencePoint
*/
public int compareDistance(Point p1, Point p2){
int distance = 0;
return distance;
}
仅此而已，在陷入困境之前，我只是在比较方法上做了一些小事情。
quickSort和分区方法是非常标准的，但我会添加它们，这样你们就可以全面了解所有内容：
/**
* Sort the array points[] in the increasing order of polar angle with respect to lowestPoint.
* Use quickSort. Construct an object of the pointComparator class with lowestPoint as the
* argument for point comparison.
*
* Ought to be private, but is made public for testing convenience.
*/
public void quickSort(){
// TODO
}
/**
* Operates on the subarray of points[] with indices between first and last.
*
* @param first starting index of the subarray
* @param last ending index of the subarray
*/
private void quickSortRec(int first, int last){
// TODO
}
/**
* Operates on the subarray of points[] with indices between first and last.
*
* @param first
* @param last
* @return
*/
private int partition(int first, int last){
// TODO
return 0;
}
我知道在快速排序方法开始之前，我基本上需要启动并运行Compare类，但我觉得我甚至不知道如何使用点/交叉比较，所以我真的很失落。
如果有人愿意帮忙，我将不胜感激！非常感谢您的光临，祝您晚上愉快。
在所有这些方法中，当您需要查看两个Point对象是否相等时，您应该使用Point的equals方法，而不是"=="：
if(p1.equals(p2)) {
//code
}
实施比较
请注意，您的compare方法需要在其实现中使用equals（）、comparePoleAngle（）和compareDistance（）。最后一组条件（返回1）也可以在else语句中处理。
public int compare(Point p1, Point p2) {
if(p1.equals(p2)) {
return 0;
}
else if(p1.equals(referencePoint) ||
(!p1.equals(referencePoint) && !p2.equals(referencePoint) && comparePolarAngle(p1, p2) == -1) ||
(!p1.equals(referencePoint) && !p2.equals(referencePoint) && comparePolarAngle(p1, p2) == 0 && compareDistance(p1, p2) == -1))
{
return -1;
}
else {
return 1;
}
}
实施compareDistance
这里我们需要的主要信息是如何仅使用点积来确定从referencePoint到Point对象的向量的长度。首先，让我们实现一个助手方法，该方法将两点作为输入，并将点积作为整数值返回。
private int dotProduct(Point p1, Point p2) {
int p1X = p1.getX() - referencePoint.getX();
int p1Y = p1.getY() - referencePoint.getY();
int p2X = p2.getX() - referencePoint.getX();
int p2Y = p2.getY() - referencePoint.getY();
//compensate for a reference point other than (0, 0)
return (p1X * p2X) + (p1Y * p2Y); //formula for dot product
}
那么，我们如何用它来计算向量的长度呢？如果我们取一个点的点积，我们得到（xx）+（yy），这是勾股定理（a^2+b^2=c^2）的左侧。因此，如果我们调用dotProduct（p1，p1），我们将得到其向量的长度平方。现在让我们实现compareDistance。
public int compareDistance(Point p1, Point p2) {
if(dotProduct(p1, p1) == dotProduct(p2, p2)) {
return 0;
}
else if(dotProduct(p1, p1) < dotProduct(p2, p2)) {
return -1;
}
else {
return 1;
}
}
不需要取点积的平方根，只需比较平方长度即可。还要注意，这里可以使用"=="，因为我们比较的是整数，而不是点。
实现comparePolaAngle
与点积一样，让我们实现一个辅助方法，用于计算两个输入点的叉积。
private int crossProduct(Point p1, Point p2) {
int p1X = p1.getX() - referencePoint.getX();
int p1Y = p1.getY() - referencePoint.getY();
int p2X = p2.getX() - referencePoint.getX();
int p2Y = p2.getY() - referencePoint.getY();
//compensate for a reference point other than (0, 0)
return (p1X * p2Y) - (p2X * p1Y); //formula for cross product
}
另一种写取两点叉积的结果的方法是|p1||p2|sin（theta），其中|p1|是p1向量的长度，|p2|是p2向量的长度并且theta是从p1到p2的角度。
相对于参考点具有相同极角的两个点的θ值为零。sin（0）=0，因此具有相同极角的两个点的叉积为零。
如果p1相对于参考点的极角小于p2的极角，则从p1到p2的角度为正。对于0<θ<180，sin（θ）为正。因此，如果我们取p1和p2的叉积，它是正的，那么p1的极角一定小于p2的极角。
如果p1相对于参考点的极角大于p2的极角，则从p1到p2的角度将是负的。对于-180<θ<0，sin（θ）为负。因此，如果我们取p1和p2的叉积，它是负的，那么p1的极角一定大于p2的极角。
使用这些信息，我们最终可以实现comparePolalAngle。
public int comparePolarAngle(Point p1, Point p2) {
if(crossProduct(p1, p2) == 0) {
return 0;
}
else if(p1.equals(referencePoint) || crossProduct(p1, p2) > 0) {
return -1;
}
else {
return 1;
}
}
我将把快速排序的实现留给您，因为我不知道您的Point对象是如何存储、访问和比较的。