Description
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
Example 1:1
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7Input: 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 2:1
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3Input: points = [[3,3],[5,-1],[-2,4]], K = 2
Output: [[3,3],[-2,4]]
(The answer [[-2,4],[3,3]] would also be accepted.)
Solution
O(NlogK), Heap1
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24class Solution {
public int[][] kClosest(int[][] points, int K) {
int[][] res = new int[K][2];
if (points.length <= K) return points;
PriorityQueue<int[]> pq = new PriorityQueue<>((a,b) -> getDistance(b) - getDistance(a));
for (int[] point: points){
pq.offer(point);
if (pq.size() > K) pq.poll();
}
for (int i = 0; i < K; i++){
int[] point = pq.poll();
res[i][0] = point[0];
res[i][1] = point[1];
}
return res;
}
public int getDistance(int[] points) {
return points[0] * points[0] + points[1] * points[1];
}
}
O(N), Divide and Conquer1
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31class Solution {
public int[][] kClosest(int[][] points, int K) {
int len = points.length, l = 0, r = len - 1;
while (l <= r) {
int mid = helper(points, l, r);
if (mid == K) break;
if (mid < K) {
l = mid + 1;
} else {
r = mid - 1;
}
}
return Arrays.copyOfRange(points, 0, K);
}
private int helper(int[][] A, int l, int r) {
int[] pivot = A[l];
while (l < r) {
while (l < r && compare(A[r], pivot) >= 0) r--;
A[l] = A[r];
while (l < r && compare(A[l], pivot) <= 0) l++;
A[r] = A[l];
}
A[l] = pivot;
return l;
}
private int compare(int[] p1, int[] p2) {
return p1[0] * p1[0] + p1[1] * p1[1] - p2[0] * p2[0] - p2[1] * p2[1];
}
}