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

我們有一個由平面上的點組成的列表 points。需要從中找出 K 個距離原點 (0, 0) 最近的點。

（這裏，平面上兩點之間的距離是歐幾裏德距離。）

你可以按任何順序返回答案。除了點坐標的順序之外，答案確保是唯一的。

示例 1：

輸入：points = [[1,3],[-2,2]], K = 1
輸出：[[-2,2]]
解釋： 
(1, 3) 和原點之間的距離為 sqrt(10)，
(-2, 2) 和原點之間的距離為 sqrt(8)，
由於 sqrt(8) < sqrt(10)，(-2, 2) 離原點更近。
我們只需要距離原點最近的 K = 1 個點，所以答案就是 [[-2,2]]。

示例 2：

輸入：points = [[3,3],[5,-1],[-2,4]], K = 2
輸出：[[3,3],[-2,4]]
（答案 [[-2,4],[3,3]] 也會被接受。） 

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 }

[Swift Weekly Contest 118]LeetCode973. 最接近原點的 K 個點 | K Closest Points to Origin