题目描述

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

代码

想到的最直接的方法，就是根据坐标的平方和排序，用一个map记录结果，key为每个点的平方和，value是结果为key的一组点。

public int[][] kClosest(int[][] points, int K) {
    if (points.length < K) return null;
    int[][] ans = new int[K][2];
    Map<Long, List<Point>> map = new HashMap<>();
    List<Long> sort = new ArrayList<>();
    for (int[] point : points) {
        long d = point[0] * point[0] + point[1] * point[1];
        if (map.containsKey(d)) {
            map.get(d).add(new Point(point));
        } else {
            ArrayList<Point> list = new ArrayList<Point>();
            list.add(new Point(point));
            map.put(d, list);
            sort.add(d);
        }
    }

    Collections.sort(sort);
    int count = 0, i = 0;
    while (count < K) {
        long d = sort.get(i++);
        List<Point> tmp = map.get(d);
        for (int i1 = 0; i1 < tmp.size(); i1++) {
            ans[count++] = new int[]{tmp.get(i1).getX(), tmp.get(i1).getY()};
        }
    }

    return ans;
}

private static class Point {
    private int x;
    private int y;

    public Point(int[] p) {
        this.x = p[0];
        this.y = p[1];
    }

    public int getX() {
        return x;
    }

    public int getY() {
        return y;
    }
}