612. K个最近的点[LintCode]

题目

给定一些 points 和一个 origin，从 points 中找到 k 个离 origin 最近的点。按照距离由小到大返回。如果两个点有相同距离，则按照x值来排序；若x值也相同，就再按照y值排序。

样例

给出
points = [[4,6],[4,7],[4,4],[2,5],[1,1]], origin = [0, 0], k = 3
返回 [[1,1],[2,5],[4,4]]

代码

/** * 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 {    public static void main(String[] args) {        int k = 3;        Point[] result = new Point[k];        Point origin = new Point(0, 0);        Point[] points = {                new Point(4, 6),                 new Point(4, 7),                 new Point(4, 4),                new Point(2, 5),                 new Point(1, 1)         };        result = kClosest(points, origin, k);        for (Point p : result) {            System.out.println("[" + p.x + "," + p.y + "]");        }    }     public Point[] kClosest(Point[] points, Point origin, int k) {        Point[] result = new Point[k];        Point tmpPoint = new Point();        for (int i = 0; i < points.length; i++) {            for (int j = i; j < points.length; j++) {                //比较点和origin点的距离,若前面的点的距离大于后面点的距离,则交换两个点的位置                if (calcLength(points[i], origin) > calcLength(points[j],                        origin)) {                    tmpPoint = points[i];                    points[i] = points[j];                    points[j] = tmpPoint;                }//如果两个点有相同距离,则按照x值来排序                else if (calcLength(points[i], origin) == calcLength(                        points[j], origin)) {                    if (points[i].x > points[j].x) {                        tmpPoint = points[i];                        points[i] = points[j];                        points[j] = tmpPoint;                    }//若x值相同，按照y值排序。                     else if (points[i].x == points[j].x) {                        if (points[i].y == points[j].y) {                            tmpPoint = points[i];                            points[i] = points[j];                            points[j] = tmpPoint;                        }                    }                }            }        }        //输出符合条件的前k的点        for (int i = 0; i < k; i++) {            result[i] = points[i];        }        return result;    }    //计算两点间距离    public static double calcLength(Point point, Point origin) {        double length = Math.sqrt((point.x - origin.x) * (point.x - origin.x)                + (point.y - origin.y) * (point.y - origin.y));        return length;    }}