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

这道题给了平面上的一系列的点，让求最接近原点的K个点。基本上没有什么难度，无非就是要知道点与点之间的距离该如何求。一种比较直接的方法就是给这个二维数组排序，自定义排序方法，按照离原点的距离从小到大排序，注意这里我们并不需要求出具体的距离值，只要知道互相的大小关系即可，所以并不需要开方。排好序之后，返回前k个点即可，参见代码如下：

解法一：

class Solution {
public:
    vector<vector<int>> kClosest(vector<vector<int>>& points, int K) {
        sort(points.begin(), points.end(), [](vector<int>& a, vector<int>& b) {
        	return a[0] * a[0] + a[1] * a[1] < b[0] * b[0] + b[1] * b[1];
        });
        return vector<vector<int>>(points.begin(), points.begin() + K);
    }
};

下面这种解法是使用最大堆 Max Heap 来做的，在 C++ 中就是用优先队列来做，这里维护一个大小为k的最大堆，里面放一个 pair 对儿，由距离原点的距离，和该点在原数组中的下标组成，这样优先队列就可以按照到原点的距离排队了，距离大的就在队首。这样每当个数超过k个了之后，就将队首的元素移除即可，最后把剩下的k个点存入结果 res 中即可，参见代码如下：

解法二：

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

Github 同步地址:

#973

类似题目：

Kth Largest Element in an Array

Top K Frequent Elements

Top K Frequent Words

参考资料：

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

https://leetcode.com/problems/k-closest-points-to-origin/discuss/217999/JavaC%2B%2BPython-O(N)

https://leetcode.com/problems/k-closest-points-to-origin/discuss/221532/C%2B%2B-STL-quickselect-priority_queue-and-multiset

https://leetcode.com/problems/k-closest-points-to-origin/discuss/220235/Java-Three-solutions-to-this-classical-K-th-problem.

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