LeetCode 973. K Closest Points to Origin

原题链接在这里：https://leetcode.com/problems/k-closest-points-to-origin/

题目：

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

题解：

Could use maxHeap to track maximum distance, keep adding, when size > K, poll.

Time Complexity: O(nlogK). n = points.length.

Space: O(K).

AC Java:

class Solution {
    public int[][] kClosest(int[][] points, int k) {
        if(points == null || points.length <= k){
            return points;
        }
        
        if(k <= 0){
            return new int[0][2];
        }
        
        PriorityQueue<int []> maxHeap = new PriorityQueue<>((a, b) -> b[0] * b[0] + b[1] * b[1] - a[0] * a[0] - a[1] * a[1]);
        for(int [] p : points){
            maxHeap.add(p);
            if(maxHeap.size() > k){
                maxHeap.poll();
            }
        }
        
        return maxHeap.toArray(new int[maxHeap.size()][2]);
    }
}

AC C++:

class Solution {
public:
    vector<vector<int>> kClosest(vector<vector<int>>& points, int k) {
        priority_queue<vector<int>> maxHeap;
        for(auto& p : points){
            int x = p[0];
            int y = p[1];
            maxHeap.push({x * x + y * y, x, y});
            if(maxHeap.size() > k){
                maxHeap.pop();
            }
        }

        vector<vector<int>> res(k);
        for(int i = 0; i < k; i++){
            vector<int> top = maxHeap.top();
            maxHeap.pop();
            res[i] = {top[1], top[2]};
        }

        return res;
    }
};

AC Python:

class Solution:
    def kClosest(self, points: List[List[int]], k: int) -> List[List[int]]:
        maxHeap = []
        for x, y in points:
            heappush(maxHeap, [-x * x - y * y, x, y])
            if len(maxHeap) > k:
                heappop(maxHeap)
        
        return [[x, y] for _, x, y in maxHeap]

类似Top K Frequent Elements.