[Swift Weekly Contest 118]LeetCode973. 最接近原点的 K 个点 | K Closest Points to Origin

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

（这里，平面上两点之间的距离是欧几里德距离。）

你可以按任何顺序返回答案。除了点坐标的顺序之外，答案确保是唯一的。

示例 1：

输入：points = [[1,3],[-2,2]], K = 1
输出：[[-2,2]]
解释：
(1, 3) 和原点之间的距离为 sqrt(10)，
(-2, 2) 和原点之间的距离为 sqrt(8)，
由于 sqrt(8) < sqrt(10)，(-2, 2) 离原点更近。
我们只需要距离原点最近的 K = 1 个点，所以答案就是 [[-2,2]]。

示例 2：

输入：points = [[3,3],[5,-1],[-2,4]], K = 2
输出：[[3,3],[-2,4]]
（答案 [[-2,4],[3,3]] 也会被接受。） 

提示：

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

 1 class Solution {
 2     func kClosest(_ points: [[Int]], _ K: Int) -> [[Int]] {
 3         var points = points
 4         points.sort(by:sortArray)
 5         var ret:[[Int]] = [[Int]]();
 6         for i in 0..<K
 7         {
 8             ret.append(points[i])
 9         }
10         return ret
11     }
12
13     func sortArray(_ a:[Int],_ b:[Int]) -> Bool
14     {
15         var da:Int = a[0]*a[0]+a[1]*a[1]
16         var db:Int = b[0]*b[0]+b[1]*b[1]
17         return (da - db) < 0
18     }
19 }

原文地址：https://www.cnblogs.com/strengthen/p/10262303.html