LeetCode #973 - K Closest Points to Origin

最新推荐文章于 2020-10-24 23:33:59 发布

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本文介绍了一种使用优先队列和自定义比较器的算法，该算法可以找出平面上距离原点最近的K个点。通过实例展示了算法的具体应用，如输入点集和K值后，如何返回最近的K个点。

题目描述：

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

class comp{
public:    
    bool operator()(vector<int> a, vector<int> b)
    { // 当返回值为true时，b在堆顶
        if((a[0]*a[0]+a[1]*a[1])<(b[0]*b[0]+b[1]*b[1])) return true;
        else return false;
    }
};

class Solution {
public:
    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
        priority_queue<vector<int>, vector<vector<int>>, comp> q;
        for(auto point:points)
        {
            q.push(point);
            if(q.size()>K) q.pop();
        }
        vector<vector<int>> result;
        while(q.size()>0)
        {
            result.push_back(q.top());
            q.pop();
        }
        return result;
    }
};