973. K Closest Points to Origin**

973. K Closest Points to Origin**

https://leetcode.com/problems/k-closest-points-to-origin/

Title Description

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

C ++ implementation 1

Quick thinking is: the smallest number you can use to retain the maximum heap, while retaining the maximum number you can use the minimum heap.

This problem requires a minimum distance point, i.e., the minimum required distance, the maximum heap. pairStored distance value and the index point, because the final point coordinates are returned.

class Solution {
private:
    inline int distance(const vector<int> &p) {
        return p[0] * p[0] + p[1] * p[1]; 
    }
public:
    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
        priority_queue<pair<int, int>> q;
        for (int i = 0; i < points.size(); ++ i) {
            q.push(std::make_pair(distance(points[i]), i));
            if (q.size() > K) q.pop();
        }
        vector<vector<int>> res;
        while (!q.empty()) {
            auto i = q.top().second;
            res.push_back(points[i]);
            q.pop();
        }
        return res;
    }
};