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 <= K <= points.length <= 10000-10000 < points[i][0] < 10000-10000 < points[i][1] < 10000
Solution
It's similar to find K largest (or smallest) elements in array.
Lazy way: Sort then pick first K elements, O(nlogn + k).
Optimized way: Use Heap, O(n + klogn), n for build heap, klogn for finding first K elements.
class Solution {
func kClosest(_ points: [[Int]], _ K: Int) -> [[Int]] {
let sorted = points.map { Point($0) } .sorted(by: <)
return sorted[..<K].map { [$0.x, $0.y] }
}
}
struct Point {
var x: Int
var y: Int
let distance: Int
init(_ data: [Int]) {
self.x = data[0]
self.y = data[1]
distance = x * x + y * y
}
}
extension Point: Comparable {
static func <(lhs: Point, rhs: Point) -> Bool {
return lhs.distance < rhs.distance
}
static func ==(lhs: Point, rhs: Point) -> Bool {
return lhs.distance == rhs.distance
}
}