目前正与Graham's Scan一起与Convex HUll合作。我是一名学生，所以我试图自己完成它，但是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，一个是随机生成的，所以我可以创建一个点数组。下一步是实现快速排序，按极坐标角度排序。这是通过比较器类完成的。比较器类是我被卡住的地方，我们被告知使用点比较和交叉比较来进行角度的比较，但我很遗憾。
/**
* 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;
}
这就是全部，我只是在比较方法上经历了一些小事情才被卡住。
quickSort和分区方法非常标准，但我会添加它们，以便你们可以广泛地看待所有内容：
/**
* 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;
}
我知道我基本上需要在启动quicksort方法之前启动并运行Compare类，但我觉得我甚至不知道如何使用点/交叉比较，所以感觉真的很丢失
如果有人愿意帮忙，我将非常感激！ 非常感谢您的光临，祝您度过愉快的夜晚。
答案 0 :(得分：2)
在所有这些方法中，当你需要查看两个Point对象是否相等时，你应该使用Point的equals方法，而不是“==”：
if(p1.equals(p2)) {
//code
}
请注意，您的比较方法需要在其实现中使用equals（），comparePolarAngle（）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;
}
}
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对象的向量长度。首先，让我们实现一个辅助方法，它将两个Point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
}
那么我们如何使用它来计算向量的长度？如果我们得到一个Point的点积，我们得到（x x）+（y y），这是毕达哥拉斯定理的左边（a ^ 2 + b ^ 2 = c ^ 2）。因此，如果我们调用dotProduct（p1，p1），我们将得到其向量的平方长度。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。另请注意，“==”可以在这里使用，因为我们比较的是整数，而不是点。
与dot产品一样，让我们实现一个帮助方法来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
}
写出两点交叉积的结果的另一种方法是| p1 || p2 | sin（theta）其中| p1 |是p1向量的长度，| p2 |是p2矢量的长度，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具有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＆lt; theta＆lt; 180，罪（theta）是正面的。因此，如果我们取p1和p2的交叉积并且它是正的，则p1的极角必须小于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的角度将为负。为-180＆lt; theta＆lt; 0，sin（theta）是负数。因此，如果我们取p1和p2的交叉积并且它是负的，则p1的极角必须大于p2的极角。
使用此信息，我们最终可以实现comparePolarAngle。
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。