[LintCode] K Closest Points

沈俭发布于2019-08-15 13:35 / 650人阅读

Problem

Given some points and a point origin in two dimensional space, find k points out of the some points which are nearest to origin.
Return these points sorted by distance, if they are same with distance, sorted by x-axis, otherwise sorted by y-axis.

Example

Given points = [[4,6],[4,7],[4,4],[2,5],[1,1]], origin = [0, 0], k = 3
return [[1,1],[2,5],[4,4]]

Solution

/**
 * Definition for a point.
 * class Point {
 *     int x;
 *     int y;
 *     Point() { x = 0; y = 0; }
 *     Point(int a, int b) { x = a; y = b; }
 * }
 */


public class Solution {
    /*
     * @param points: a list of points
     * @param origin: a point
     * @param k: An integer
     * @return: the k closest points
     */
    public Point[] kClosest(Point[] points, Point origin, int k) {
        Comparator pointComparator = new Comparator() {
            public int compare(Point A, Point B) {
                if (distance(B, origin) == distance(A, origin)) {
                    if (A.x == B.x) {
                        return A.y - B.y;
                    } else {
                        return A.x - B.x;
                    }
                } else {
                    //maintain a min heap
                    return distance(A, origin) > distance(B, origin) ? 1 : -1;
                }
            }
        };
        
        PriorityQueue Q = new PriorityQueue(k, pointComparator);
        
        for (Point point: points) {
            Q.add(point);
        }
        Point[] res = new Point[k];
        for (int i = 0; i < k; i++) {
            res[i] = Q.poll();
        }
        return res;
    }
    
    public double distance(Point A, Point B) {
        int l = Math.abs(A.x - B.x);
        int w = Math.abs(A.y - B.y);
        return Math.sqrt(Math.pow(l, 2) + Math.pow(w, 2));
    }
}

云服务器 GPU云服务器 Points LintCode closest() closest

文章版权归作者所有，未经允许请勿转载,若此文章存在违规行为，您可以联系管理员删除。

转载请注明本文地址：https://www.ucloud.cn/yun/68189.html

Leetcode PHP题解--D29 973. K Closest Points to Origi

摘要：题目链接题目分析给一个坐标数组，从中返回个离最近的坐标。其中，用欧几里得距离计算。思路把距离作为数组的键，把对应坐标作为数组的值。用函数排序，再用函数获取前个即可。最终代码若觉得本文章对你有用，欢迎用爱发电资助。 973. K Closest Points to Origin 题目链接 973. K Closest Points to Origin 题目分析给一个坐标数组points...

Sanchi 2019-07-01 12:39 评论0 收藏0
[LintCode/LeetCode] 3Sum Closest

摘要：这个题和的做法基本一样，只要在循环内计算和最接近的和，并赋值更新返回值即可。 Problem Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three intege...

ShevaKuilin 2019-08-14 15:42 评论0 收藏0
[LintCode] Max Points on a Line

摘要：两次循环对中第个和第个进行比较设置重复点数，相同斜率点数。内部循环每次结束后更新和点相同斜率的点的最大数目。外部循环每次更新为之前结果和本次循环所得的较大值。重点一以一个点为参照求其他点连线的斜率，不需要计算斜率。 Problem Given n points on a 2D plane, find the maximum number of points that lie on th...

bingo 2019-08-14 15:14 评论0 收藏0
[LintCode] Baseball Game

Problem Youre now a baseball game point recorder. Given a list of strings, each string can be one of the 4 following types: Integer (one rounds score): Directly represents the number of points you get...

newtrek 2019-08-16 12:36 评论0 收藏0