K个最近的点

题目描述：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。

Example

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]]

思路：调用STL自带算法void sort(RandomAccessIterator beg,RandomAccessIterator end,BinaryPredicate op)对结构体struct Point进行排序，

      其中最重要的怎样设计op这个函数；op函数主要根据题目描述设计，完整代码如下：

#include <iostream>#include<algorithm>using namespace std;struct Point{    int x;    int y;    Point():x(0),y(0) {}    Point(int a,int b) : x(a),y(b) {}};static Point c(0,0);//设置全局变量，用来存储originclass Solution{public:    vector<Point> kClosest(vector<Point> points,Point origin,int k);};double distance(Point a,Point b){   return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));}bool less_distance(Point a,Point b)//op设计{    double d1 = distance(a,c);    double d2 = distance(b,c);    if(d1 <  d2)        return true;    else if(d1 == d2)    {        if(a.x < b.x)            return true;        else if(a.x == b.x)        {            if(a.y < b.y)                return true;            else                return false;        }        else            return false;    }    else        return false;}vector<Point> Solution::kClosest(vector<Point> points,Point origin,int k){        c.x = origin.x;        c.y = origin.y;//更新origin点坐标        vector<Point> Result;        Result.clear();        sort(points.begin(),points.end(),less_distance);        for(int i=0;i<k;i++)        {           Result.push_back(points[i]);        }       return Result;}int main()//测试代码{   vector<Point> points = {Point(4,6),Point(4,7),Point(4,4),Point(2,5),Point(1,1)};   Point c(0,0);   Solution m;   vector<Point> Result;   Result = m.kClosest(points,c,3);   for(int i=0;i<3;i++)   {       cout << Result[i].x << "," << Result[i].y << endl;   }    cout << endl;    return 0;}