所以我尝试写一个简单的方法，需要在设置中的四个坐标，并决定他们是否形成一个正方形或not.My的做法是先从一个点，计算其他三个点和基础点之间的距离。由此我们可以得到双方具有相同的值，其中一个是diagonal.Then我用毕达哥拉斯定理发现，如果双方平方等于diagonal.If这是isSquare方法返回TRUE，否则false.The事我想找出是有一些情况下，我可能会错过了或者如果事情是错的approach.Thanks的所有帮助。

 公共类CoordinatesSquare {

公共静态布尔isSquare（名单＆LT;点＆GT; listPoints）{
    如果（listPoints = NULL和放大器;！＆安培; listPoints.size（）== 4）{
        INT距离1 =距离（listPoints.get（0），listPoints.get（1））;
        INT distance2的距离=（listPoints.get（0），listPoints.get（2））;
        INT距离3 =距离（listPoints.get（0），listPoints.get（3））;

        如果（距离1 ==距离2）{
            //检查是否边等于对角线
            如果（距离3 ==距离1 +距离2）{
                返回true;
            }

        }否则，如果（距离1 ==距离3）{
            //检查是否边等于对角线
            如果（距离2 ==距离1 +距离3）{
                返回true;
            }

        }
    }
    返回false;
}

私有静态诠释距离（点对点，点点2）{
              //（X2-X1）^ 2 +（Y2-Y1）^ 2
    返回（int）的（Math.pow（point2.x  -  point.x，2）+（Mat​​h.pow（point2.y
             -  point.y，2）））;
}

公共静态无效的主要（字符串的args []）{
    名单＆LT;点＆GT; pointz =新的ArrayList＆LT;点＆GT;（）;
    pointz.add（新点（2,2））;
    pointz.add（新点（2,4））;
    pointz.add（新点（4,2））;
    pointz.add（新点（4,4））;
    的System.out.println（CoordinatesSquare.isSquare（pointz））;
}
}


//点​​类
 公共类点{
整数x;
码整数Y;
布尔isVisited;

公共点（范围整数x，整数Y）{
    this.x = X;
    this.y = Y;
}

@覆盖
公共布尔等于（obj对象）{
    如果（物镜=空＆安培;！＆安培; obj.getClass（）等于（this.getClass（））。）{
        返程（（点）OBJ）.x.equals（this.x）及及（（点）OBJ）.y.equals（this.y）;
    }
    返回false;

}
}
 

解决方案

您知道吗，你可以做同样的检查更容易。你只需要检查两件事情： 四点做平行四边形和它的角度一个是正确的。

首先是真实的，当 P3 = P1 +（P2-P1）+（P4-P1）

和第二时（P2-P1）*（P4-P1）= 0

其中， A * B 是一个点积（AX * Bx的+好哦*通过）

在这里，唯一的缺点是计算错误。你不能指望花车是完全相等，所以不是 A = B 你应该考虑使用类似 ABS（AB）＆LT; Ë，其中电子是你的情况足够小。

So I am trying to write a simple method which takes in set of four coordinates and decide whether they form a square or not.My approach is start with a point and calculate the distance between the other three points and the base point.From this we can get the two sides which have same value and the one which is a diagonal.Then I use Pythagoras theorem to find if the sides square is equal to the diagonal.If it is the isSquare method return true else false.The thing I want to find out is there some cases I might be missing out on or if something is wrong with the approach.Thanks for the all the help.

public class CoordinatesSquare {

public static boolean isSquare(List<Point> listPoints) {
    if (listPoints != null && listPoints.size() == 4) {
        int distance1 = distance(listPoints.get(0), listPoints.get(1));
        int distance2 = distance(listPoints.get(0), listPoints.get(2));
        int distance3 = distance(listPoints.get(0), listPoints.get(3));

        if (distance1 == distance2) {
            // checking if the sides are equal to the diagonal
            if (distance3 == distance1 + distance2) {
                return true;
            }

        } else if (distance1 == distance3) {
            // checking if the sides are equal to the diagonal
            if (distance2 == distance1 + distance3) {
                return true;
            }

        }
    }
    return false;
}

private static int distance(Point point, Point point2) {
              //(x2-x1)^2+(y2-y1)^2
    return (int) (Math.pow(point2.x - point.x, 2) + (Math.pow(point2.y
            - point.y, 2)));
}

public static void main(String args[]) {
    List<Point> pointz = new ArrayList<Point>();
    pointz.add(new Point(2, 2));
    pointz.add(new Point(2, 4));
    pointz.add(new Point(4, 2));
    pointz.add(new Point(4, 4));
    System.out.println(CoordinatesSquare.isSquare(pointz));
}
}


//Point Class
 public class Point {
Integer x;
Integer y;
boolean isVisited;

public Point(Integer x, Integer y) {
    this.x = x;
    this.y = y;
}

@Override
public boolean equals(Object obj) {
    if(obj!=null && obj.getClass().equals(this.getClass())){
        return ((Point) obj).x.equals(this.x)&&((Point) obj).y.equals(this.y);
    }
    return false;

}
}

You know, you can do the same check much easier. You just have to check two things: "four points make a parallelogram" and "one of its angles is right".

First is true when P3 = P1 + (P2-P1) + (P4-P1)

And the second when (P2-P1)*(P4-P1) = 0

Where A*B is a dot product (A.x * B.x + A.y * B.y)

The only catch here is computational error. You can't expect floats to be exactly equal, so instead of A=B you should consider using something like abs(A-B) < E where E is small enough for your case.