在C ++中找到K个最接近原点的点
假设我们有一组要点。我们的任务是找到最接近原点的K个点。假设点是(3,3)，(5,-1)和(-2,4)。然后最接近的两个（K=2）点是(3,3)，(-2,4)。
为了解决这个问题，我们将基于点的欧几里得距离对列表进行排序，然后从排序的列表中获取最上面的K个元素。这些是K最近的点。
示例
#include<iostream>
#include<algorithm>
using namespace std;
class Point{
private:
int x, y;
public:
Point(int x = 0, int y = 0){
this->x = x;
this->y = y;
}
void display(){
cout << "("<<x<<", "<<y<<")";
}
friend bool comparePoints(Point &p1, Point &p2);
};
bool comparePoints(Point &p1, Point &p2){
float dist1 = (p1.x * p1.x) + (p1.y * p1.y);
float dist2 = (p2.x * p2.x) + (p2.y * p2.y);
return dist1 < dist2;
}
void closestKPoints(Point points[], int n, int k){
sort(points, points+n, comparePoints);
for(int i = 0; i<k; i++){
points[i].display();
cout << endl;
}
}
int main() {
Point points[] = {{3, 3},{5, -1},{-2, 4}};
int n = sizeof(points)/sizeof(points[0]);
int k = 2;
closestKPoints(points, n, k);
}输出结果
(3, 3) (-2, 4)