[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,[5,-1],4]],K = 2 Output: [[3,4]] (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

我们有一个由平面上的点组成的列表 points。需要从中找出 K 个距离原点 (0,0) 最近的点。

（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,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,K = 2
输出：[[3,4]]
（答案 [[-2,3]] 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

1064ms 

 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 }