We have a list of points on the plane.  Find the K closest points to the origin (0, 0).

(Here, the distance between two points on a plane is the Euclidean distance.)

You may return the answer in any order.  The answer is guaranteed to be unique (except for the order that it is in.)

Example 1:

Input: points = [[1,3],[-2,2]], K = 1
Output: [[-2,2]]
Explanation:
The distance between (1, 3) and the origin is sqrt(10).
The distance between (-2, 2) and the origin is sqrt(8).
Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin.
We only want the closest K = 1 points from the origin, so the answer is just [[-2,2]].

Example 2:

Input: points = [[3,3],[5,-1],[-2,4]], K = 2
Output: [[3,3],[-2,4]]
(The answer [[-2,4],[3,3]] would also be accepted.)

Note:

  1. 1 <= K <= points.length <= 10000
  2. -10000 < points[i][0] < 10000
  3. -10000 < points[i][1] < 10000

没啥好说的,写的差劲了

struct P{
double dis;
int adr;
}H[];
bool cmd(P x,P y){
if(x.dis <= y.dis){
return ;
}
return ;
}
class Solution {
public:
vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
vector<vector<int>> x;
vector<int>y; int Len = points.size();
for(int i=;i<Len;i++){
int a = points[i][];
int b = points[i][];
//cout<<a<<" "<<b<<endl;
double result = sqrt(a*a+b*b);
H[i].dis = result;
H[i].adr = i;
}
sort(H,H+Len,cmd);
for(int i=;i<K;i++){
x.push_back(points[H[i].adr]);
}
return x;
}
};

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