# [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。）

```输入：points = [[1,3],[-2,2]], K = 1

(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)，

```

```输入：points = [[3,3],[5,-1],[-2,4]], K = 2

（答案 [[-2,4],[3,3]] 也会被接受。） ```

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 }```

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

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