查找到原点的K个最近点

给定二维平面上的点列表和整数K。任务是找到最接近原点的K个点并将其打印出来。
注意：平面上两点之间的距离是欧几里得距离。

例子：

Input : point = [[3, 3], [5, -1], [-2, 4]], K = 2
Output : [[3, 3], [-2, 4]]
Square of Distance of origin from this point is 
(3, 3) = 18
(5, -1) = 26
(-2, 4) = 20
So rhe closest two points are [3, 3], [-2, 4].

Input : point = [[1, 3], [-2, 2]], K  = 1
Output : [[-2, 2]]
Square of Distance of origin from this point is
(1, 3) = 10
(-2, 2) = 8 
So the closest point to origin is (-2, 2)

方法：想法是为每个给定点计算距原点的欧几里得距离，并根据找到的欧几里得距离对数组进行排序。打印列表中的前k个最接近的点。

算法 ：
考虑两个坐标分别为(x1，y1)和(x2，y2)的点。这两个点之间的欧式距离为：

√{(x2-x1)2 + (y2-y1)2}

1. 使用欧几里德距离公式按距离对点进行排序。
2. 从列表中选择前K个点
3. 以任何顺序打印获得的点。

下面是上述方法的实现：

C++

// C++ program for implementation of 
// above approach
#include
using namespace std;
 
// Function to print required answer
void pClosest(vector> pts, int k)
{
     
    // In multimap values gets
    // automatically sorted based on
    // their keys which is distance here
    multimap mp;
    for(int i = 0; i < pts.size(); i++)
    {
        int x = pts[i][0], y = pts[i][1];
        mp.insert({(x * x) + (y * y) , i});
    }
     
    for(auto it = mp.begin();
             it != mp.end() && k > 0;
             it++, k--)
        cout << "[" << pts[it->second][0] << ", "
             << pts[it->second][1] << "]" << "\n";
}
 
// Driver code
int main()
{
    vector> points = { { 3, 3 },
                                   { 5, -1 },
                                   { -2, 4 } };
     
    int K = 2;
     
    pClosest(points, K);
    return 0;
}
 
// This code is contributed by sarthak_eddy.

Java

// Java program for implementation of 
// above approach
import java.util.*;
 
class GFG{
     
// Function to print required answer
static void pClosest(int [][]pts, int k)
{
    int n = pts.length;
    int[] distance = new int[n];
    for(int i = 0; i < n; i++)
    {
        int x = pts[i][0], y = pts[i][1];
        distance[i] = (x * x) + (y * y);
    }
 
    Arrays.sort(distance);
     
    // Find the k-th distance
    int distk = distance[k - 1];
 
    // Print all distances which are
    // smaller than k-th distance
    for(int i = 0; i < n; i++)
    {
        int x = pts[i][0], y = pts[i][1];
        int dist = (x * x) + (y * y);
         
        if (dist <= distk)
            System.out.println("[" + x + ", " + y + "]");
    }
}
 
// Driver code
public static void main (String[] args)
{
    int points[][] = { { 3, 3 },
                       { 5, -1 },
                       { -2, 4 } };
 
    int K = 2;
     
    pClosest(points, K);
}
}
 
// This code is contributed by sarthak_eddy.

Python3

# Python3 program for implementation of
# above approach
 
# Function to return required answer
def pClosest(points, K):
 
    points.sort(key = lambda K: K[0]**2 + K[1]**2)
 
    return points[:K]
 
# Driver program
points = [[3, 3], [5, -1], [-2, 4]]
 
K = 2
 
print(pClosest(points, K))

C#

// C# program for implementation
// of above approach
using System;
class GFG{
     
// Function to print
// required answer
static void pClosest(int [,]pts,
                     int k)
{
  int n = pts.GetLength(0);
 
  int[] distance = new int[n];
   
  for(int i = 0; i < n; i++)
  {
    int x = pts[i, 0],
        y = pts[i, 1];
    distance[i] = (x * x) +
                  (y * y);
  }
 
  Array.Sort(distance);
 
  // Find the k-th distance
  int distk = distance[k - 1];
 
  // Print all distances which are
  // smaller than k-th distance
  for(int i = 0; i < n; i++)
  {
    int x = pts[i, 0],
        y = pts[i, 1];
    int dist = (x * x) +
               (y * y);
 
    if (dist <= distk)
      Console.WriteLine("[" + x +
                        ", " + y + "]");
  }
}
 
// Driver code
public static void Main (string[] args)
{
  int [,]points = {{3, 3},
                   {5, -1},
                   {-2, 4}};
  int K = 2;
  pClosest(points, K);
}
}
 
// This code is contributed by Chitranayal

输出：

[[3, 3], [-2, 4]]

复杂度分析：

• 时间复杂度： O(n log n)。
  查找每个点到原点的距离的时间复杂度为O(n)，对数组进行排序的时间复杂度为O(n log n)
• 空间复杂度： O(1)。
  由于不需要额外的空间。