[LeetCode] 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

最接近原点的K个点。

题意是给一些点的坐标（以二维数组表示）和一个数字K。求距离最靠近原点的前K个点。

求前K个XX的问题，十有八九可以用heap/priority queue解决，这个题也不例外。我这里给出pq的解法，用pq创建一个最小堆，将每个坐标与原点的距离加入pq。求距离就是初中数学的勾股定理。

时间O(nlogk)

空间O(n)

Java实现

 1 class Solution {
 2     public int[][] kClosest(int[][] points, int K) {
 3         PriorityQueue<int[]> pq = new PriorityQueue<>(K,
 4                 (o1, o2) -> o1[0] * o1[0] + o1[1] * o1[1] - o2[0] * o2[0] - o2[1] * o2[1]);
 5         for (int[] point : points) {
 6             pq.add(point);
 7         }
 8         int[][] res = new int[K][2];
 9         for (int i = 0; i < K; i++) {
10             res[i] = pq.poll();
11         }
12         return res;
13     }
14 }

这道题还有一个类似快速排序的做法，也值得掌握。

时间O(n), worst case, O(n^2)

空间O(1)

Java实现

 1 class Solution {
 2     public int[][] kClosest(int[][] points, int K) {
 3         int len = points.length;
 4         int l = 0;
 5         int r = len - 1;
 6         while (l <= r) {
 7             int mid = helper(points, l, r);
 8             if (mid == K) {
 9                 break;
10             }
11             if (mid < K) {
12                 l = mid + 1;
13             } else {
14                 r = mid - 1;
15             }
16         }
17         return Arrays.copyOfRange(points, 0, K);
18     }
19 
20     private int helper(int[][] points, int left, int right) {
21         int[] pivot = points[left];
22         while (left < right) {
23             while (left < right && compare(points[right], pivot) >= 0) {
24                 right--;
25             }
26             points[left] = points[right];
27             while (left < right && compare(points[left], pivot) <= 0) {
28                 left++;
29             }
30             points[right] = points[left];
31         }
32         points[left] = pivot;
33         return left;
34     }
35 
36     private int compare(int[] p1, int[] p2) {
37         return p1[0] * p1[0] + p1[1] * p1[1] - p2[0] * p2[0] - p2[1] * p2[1];
38     }
39 }