【发布时间】：2015-04-05 01:42:44
【问题描述】：
目前正在与 Convex HUll 一起使用 Graham's Scan。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站点以找到答案。简而言之，我有我的构造函数，一个来自文件，一个随机生成，可以工作，所以我能够创建一个点数组。下一步是实现快速排序，按极角排序。这是通过比较器类完成的。比较器类是我卡住的地方，我们被告知使用点比较和交叉比较来进行角度比较，但我很迷茫。
/**
* 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;
}
就是这样，我只是在比较方法上经历了一些小事情，然后才卡住。
quickSort 和 partition 方法是相当标准的，但我会添加它们，以便你们可以广泛了解所有内容：
/**
* 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;
}
我知道我基本上需要启动并运行 Compare 类，然后才能启动快速排序方法，但我觉得我什至不知道如何使用点/交叉比较，所以我真的很迷茫.
如果有人愿意提供帮助，我将非常感激！ 非常感谢您的观看，祝您有个愉快的夜晚。
【问题讨论】：
标签： java dot-product cross-product grahams-scan