'lintcode K个最近的点'

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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。

样例

例1:

输入: points = [[4,6],[4,7],[4,4],[2,5],[1,1]], origin = [0, 0], k = 3
输出: [[1,1],[2,5],[4,4]]
例2:

输入: points = [[0,0],[0,9]], origin = [3, 1], k = 1
输出: [[0,0]]

思考

根据到origin点的距离进行排序，取出前k个

代码

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

class Solution {
public:
    /**
     * @param points: a list of points
     * @param origin: a point
     * @param k: An integer
     * @return: the k closest points
     */
    static bool cmp(const Point A, const Point B){
        int x1 = A.x, x2 = B.x, y1 = A.y, y2 = B.y;
        if (x1 * x1 +y1 * y1 < x2 * x2 + y2* y2)
            return true;
        else if (x1 * x1 + y1 * y1 > x2 * x2 + y2 * y2)
            return false;
        else
            return x1 == x2 ? y1 < y2: x1 < x2;
    }
    vector<Point> kClosest(vector<Point> &points, Point &origin, int k) {
        // write your code here
        vector<Point> res;
        if (k == 0) return res;
        for (int i = 0; i < points.size();i++) {
            points[i].x -= origin.x;
            points[i].y -= origin.y;
        }
        sort(points.begin(), points.end(), cmp);
        for (int i = 0; i < k; i++) {
            points[i].x += origin.x;
            points[i].y += origin.y;
            res.push_back(points[i]);
        }
        return res;
    }
};
-------------end of filethanks for reading-------------