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.)
代码
Approach 1: Sort
classSolution {publicint[][] kClosest(int[][] points,int K) {int N =points.length;int[] dists =newint[N];for (int i =0; i < N; i++) { dists[i] =dist(points[i]); }Arrays.sort(dists);int distK = dists[k -1];int[][] ans =newint[K][2];int t =0;for (int i =0; i < N; i++) {if (dist(points[i])<= distK) { ans[t++] = points[i]; } }return ans; }publicintdist(int[] point) {return point[0] * point[0] + point[1] * point[1]; }}
the average time complexity is O(N) , but just like quick sort, in the worst case, this solution would be degenerated to O(N^2), and pratically, the real time it takes on leetcode is 15ms.