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 <= K <= points.length <= 10000
-10000 < points[i][0] < 10000
-10000 < points[i][1] < 10000
没啥好说的,写的差劲了
struct P{ double dis; int adr; }H[20000]; bool cmd(P x,P y){ if(x.dis <= y.dis){ return 1; } return 0; } 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=0;i<Len;i++){ int a = points[i][0]; int b = points[i][1]; //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=0;i<K;i++){ x.push_back(points[H[i].adr]); } return x; } };