@Yano 2019-01-13T07:33:01.000000Z 字数 1548 阅读 1096

# LeetCode 973 K Closest Points to Origin

LeetCode

# 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

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;    }}

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