LeetCode--No.973--K Closest Points to Origin

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

知道PriorityQueue的存在，但是被我给忘了。抄的答案代码。
今天要再看一下priorityqueue的comparator定义等等
PriorityQueue 实现的是最大堆/最小堆
在这里我们用了最大堆，虽然我们想要取K个最小距离的point, 但是用最大堆，我们才可以非常快速地删掉最大的元素，保留最小的那一个。
针对comparator函数, (p1, p2), 在这里p1是新的元素，p2是原来的元素。所以我们用p2 - p1 来表示最大堆，p1-p2来表示最小堆

最小堆的原理：当前节点与父节点比较，如果比父节点小，就交换。或者说，如果comparator函数返回为负，就交换。
在最大堆里面，我们希望，如果当前节点比父节点大， comparator返回为负，然后交换。所以用p2-p1来表示。

理解起来，我们可以认为：p1-p2 是自然顺序，代表默认的情况，例如，最小堆/升序排列。如果想要违背一下自然顺序，那么就用相反的p2-p1即可。

class Solution {
    public int[][] kClosest(int[][] points, int K) {
        PriorityQueue<int[]> pq = new PriorityQueue<int[]>((p1, p2) -> p2[0]*p2[0] + p2[1] * p2[1] - p1[0] * p1[0] - p1[1] * p1[1]);
        for(int[] p : points){
            pq.offer(p);
            if (pq.size() > K){
                pq.poll();
            }
        }
        int[][] res = new int[K][2];
        while( K > 0){
            res[--K] = pq.poll();
        }
        return res;
    }
}

今天好累，很多事情都做不完。只刷了两道题，希望明天可以加油。

LeetCode--No.973--K Closest Points to Origin

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