前往小程序,Get更优阅读体验!
立即前往
首页
学习
活动
专区
工具
TVP
发布
社区首页 >专栏 >LeetCode 973 K Closest Points to Origin

LeetCode 973 K Closest Points to Origin

作者头像
Yano_nankai
发布2019-02-25 16:26:51
4240
发布2019-02-25 16:26:51
举报
文章被收录于专栏:二进制文集二进制文集

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:

代码语言:javascript
复制
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:

代码语言:javascript
复制
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:

代码语言:javascript
复制
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。

代码语言:javascript
复制
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;
    }
}
本文参与 腾讯云自媒体分享计划,分享自作者个人站点/博客。
原始发表:2019.01.13 ,如有侵权请联系 cloudcommunity@tencent.com 删除

本文分享自 作者个人站点/博客 前往查看

如有侵权,请联系 cloudcommunity@tencent.com 删除。

本文参与 腾讯云自媒体分享计划  ,欢迎热爱写作的你一起参与!

评论
登录后参与评论
0 条评论
热度
最新
推荐阅读
目录
  • 题目描述
  • 代码
领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档