LintCode K个最近的点

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]]

坐标点类

class Point {    int x;    int y;    Point() {        x = 0;        y = 0;    }    Point(int a, int b) {        x = a;        y = b;    }    @Override    public String toString() {        return "[" + x + "," + y + "]";    }}

思路

1. 计算距离^2，获得与点等长的数组，切距离与点的坐标一一对应。
2. 循环k次，寻找k个最小距离，按照最小距离的下标，并记录最近距离点。
3. 每次比较，都要交换距离调整和点在数组中的位置，防止丢失与重复。

代码

public static Point[] kClosest(Point[] points, Point origin, int k) {        // Write your code here        int[] distance = new int[points.length];        //计算距离^2        for (int i = 0; i < points.length; i++) {            distance[i] = (points[i].x - origin.x) * (points[i].x - origin.x)             + (points[i].y - origin.y) * (points[i].y - origin.y);        }        Point[] ps = new Point[k];        //获取距离最近的k个坐标        for (int i = 0; i < k; i++) {            int min = distance[i];            ps[i] = points[i];            for (int j = i + 1; j < distance.length; j++) {                if (min > distance[j]) {                    min = distance[j];                    //交换，调整对应距离值，防止重复                    int t = distance[i];                    distance[i] = distance[j];                    distance[j] = t;                    ps[i] = points[j];                    //交换，调整对应坐标，防止重复                    Point o = points[j];                    points[j] = points[i];                    points[i] = o;                }                if (min == distance[j]) {                    if (points[i].x > points[j].x) {                        ps[i] = points[j];                        //交换，调整对应坐标，防止重复                        Point o = points[j];                        points[j] = points[i];                        points[i] = o;                    } else if (points[i].x == points[j].x) {                        if (points[i].y > points[j].y) {                            ps[i] = points[j];                            //交换，调整对应坐标，防止重复                            Point o = points[j];                            points[j] = points[i];                            points[i] = o;                        }                    }                }            }        }        return ps;    }