973. K Closest Points to Origin

问题描述：

We have a list of points on the plane.  Find the K closest points to the origin (0, 0).

(Here, the distance between two points on a plane is the Euclidean distance.)

You may return the answer in any order.  The answer is guaranteed to be unique (except for the order that it is in.)

 

Example 1:

Input: points = [[1,3],[-2,2]], K = 1
Output: [[-2,2]]
Explanation: 
The distance between (1, 3) and the origin is sqrt(10).
The distance between (-2, 2) and the origin is sqrt(8).
Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin.
We only want the closest K = 1 points from the origin, so the answer is just [[-2,2]].

Example 2:

Input: points = [[3,3],[5,-1],[-2,4]], K = 2
Output: [[3,3],[-2,4]]
(The answer [[-2,4],[3,3]] would also be accepted.)

 

Note:

1. 1 <= K <= points.length <= 10000
2. -10000 < points[i][0] < 10000
3. -10000 < points[i][1] < 10000

 

解题思路：

重写比较方法，比较距离，使用最小堆

注意值的范围来选择数据类型

 

代码：

struct cmp{
    bool operator () (vector<int> &p1, vector<int> &p2){
        int d1 = pow(p1[0], 2) + pow(p1[1], 2);
        int d2 = pow(p2[0], 2) + pow(p2[1], 2);
        return d1 > d2;
    }
};
class Solution {
public:
    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
        vector<vector<int> > ret;
        if(points.empty()) return ret; 
        priority_queue<vector<int>, vector<vector<int>> , cmp> q;
        for(auto p : points){
            q.push(p);
        }
        while(!q.empty() && K != 0){
            ret.push_back(q.top());
            q.pop();
            K--;
        }
        return ret;
    }
};