973. 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

Rust Solution

struct 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);
}

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