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] < 10000struct Solution;
use std::cmp::Ordering::*;
fn distance(v: &[i32]) -> i32 {
v[0] * v[0] + v[1] * v[1]
}
fn quick_select(a: &mut Vec<Vec<i32>>, l: usize, r: usize, k: usize) {
if l == r {
return;
}
let index = partition(a, l, r);
let rank = index - l + 1;
match rank.cmp(&k) {
Greater => quick_select(a, l, index - 1, k),
Less => quick_select(a, index + 1, r, k - rank),
_ => {}
}
}
fn partition(a: &mut Vec<Vec<i32>>, l: usize, r: usize) -> usize {
let x = distance(&a[r]);
let mut i = l;
for j in l..r {
if distance(&a[j]) <= x {
a.swap(i, j);
i += 1;
}
}
a.swap(i, r);
i
}
impl Solution {
fn k_closest(mut points: Vec<Vec<i32>>, k: i32) -> Vec<Vec<i32>> {
let n = points.len();
quick_select(&mut points, 0, n - 1, k as usize);
points.resize(k as usize, vec![]);
points
}
}
#[test]
fn test() {
let points: Vec<Vec<i32>> = vec_vec_i32![[1, 3], [-2, 2]];
let res: Vec<Vec<i32>> = vec_vec_i32![[-2, 2]];
assert_eq!(Solution::k_closest(points, 1), res);
}