973. K Closest Points to Origin --------M

 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

A:

就是用heap   ,，  选择了max heap, 这样，可以只用 K个。  然后，每次加入时候，对比头部

class Solution {
public:
    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
        priority_queue<vector<int>, vector<vector<int>>, comparator> maxHeap;
        for(auto & p : points){
            if(maxHeap.size()==K ){
                auto & t = maxHeap.top();
                if(t[0]*t[0] + t[1]*t[1] > p[0]*p[0] + p[1]*p[1]){
                    maxHeap.pop();
                    maxHeap.push(p);
                }
            }else{
                maxHeap.push(p);
            }
        }
        vector<vector<int>> res;
        while(not maxHeap.empty()){
            auto t = maxHeap.top();
            maxHeap.pop();
            res.push_back(t);
        }
        return res;
    }
private:
    struct comparator{
        bool operator()(const vector<int> & a, const vector<int> &b){
            return a[0]*a[0] + a[1]*a[1] < b[0]*b[0] + b[1]*b[1];
        }
    };
};

Mistakes：

auto & t = maxHeap.top();
maxheap.pop();

一开始，我写成这样，想重复 利用这个空间来着。    然而 t  可能已经被销毁了。会有heap error