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📜  在另一个点旋转180度时反射一个点

📅  最后修改于: 2021-04-29 16:59:38             🧑  作者: Mango

给定两个点，二维平面上的坐标(x 1 ，y 1 )和(x 2 ，y 2 ) 。任务是找到在(x 2 ，y 2 )旋转180度时(x 1 ，y 1 )的反射。
例子： 

Input : x1 = 0, y1 = 0, x2 = 1, y2 = 1
Output : (2, 2)

Input : x1 = 1, y1 = 1, x2 = 2, y2 = 2
Output : (3, 3)

令关于(x 2 ，y 2 )的点(x 1 ，y 1 )的反射点为(x’，y’)。
对于(x’，y’)是点(x 1 ，y 1 )围绕点(x 2 ，y 2 )旋转180度，它们必须全部共线，即所有三个点必须位于同一条直线上。同样，观察(x 2 ，y 2 )将成为(x 1 ，y 1 )与(x’，y’)之间的中点。

所以，
x’– x 2 = x 2 – x 1
y’– y 2 = y 2 – y 1
x’= 2 * x 2 – x 1
y’= 2 * y 2 – y 1
以下是此方法的实现：

C++
// CPP Program tof find the 180 degree reflection
// of one point around another point.
#include 
using namespace std;
 
void findPoint(int x1, int y1, int x2, int y2)
{
    cout << "(" << 2 * x2 - x1 << ", "
         << 2 * y2 - y1 << ")";
}
 
 
int main()
{
    int x1 = 0, y1 = 0, x2 = 1, y2 = 1;
    findPoint(x1, y1, x2, y2);
    return 0;
}

Java
// Java Program to find the 180 degree
// reflection of one point around
// another point.
class GFG {
     
    static void findPoint(int x1, int y1,
                          int x2, int y2)
    {
        System.out.println("(" + (int)(2 * x2 - x1)
               + "," + (int)(2 * y2 - y1 ) + " )");
    }
     
    // Driver code
    public static void main(String args[])
    {
        int x1 = 0, y1 = 0, x2 = 1, y2 = 1;
         
        findPoint(x1, y1, x2, y2);
    }
}
 
// This code is contributed by Arnab Kundu.

Python3
# Python3 Program for find the 180
# degree reflection of one point
# around another point.
 
def findPoint(x1, y1, x2, y2):
    print("(" , 2 * x2 - x1 , ",",
                2 * y2 - y1 ,")");
 
# Driver Code
x1 = 0;
y1 = 0;
x2 = 1;
y2 = 1;
findPoint(x1, y1, x2, y2);
 
# This code is contributed by mits

C#
// C# Program to find the 180 degree reflection
// of one point around another point.
using System;
 
public class GFG {
     
    static void findPoint(int x1, int y1,
                          int x2, int y2)
    {
        Console.WriteLine("(" + (int)(2 * x2 - x1)
               + "," + (int)(2 * y2 - y1 ) + " )");
    }
     
    // Driver code
    static public void Main(String []args)
    {
        int x1 = 0, y1 = 0, x2 = 1, y2 = 1;
         
        findPoint(x1, y1, x2, y2);
    }
}
 
// This code is contributed by Arnab Kundu.

PHP

Javascript

输出： 
(2, 2)

时间复杂度： O(1)