973. K Closest Points to Origin - Medium
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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
M1: min heap
time = O(n + klogn), space = O(n)
class Solution public int[][] kClosest(int[][] points, int K) int[][] res = new int[K][2]; if(points == null || points.length == 0 || K == 0) return res; PriorityQueue<int[]> minHeap = new PriorityQueue<>(new Comparator<int[]>() @Override public int compare(int[] p1, int[] p2) double d1 = getDistance(p1, new int[] 0, 0); double d2 = getDistance(p2, new int[] 0, 0); if(d1 == d2) return 0; return d1 < d2 ? -1 : 1; ); for(int[] point : points) minHeap.offer(point); for(int i = 0; i < K; i++) res[i] = minHeap.poll(); return res; public double getDistance(int[] p, int[] o) int x = p[0] - o[0]; int y = p[1] - o[1]; return Math.sqrt(x * x + y * y);
M2: max heap
time = O(k + (n-k)logk), space = O(k)
class Solution public int[][] kClosest(int[][] points, int K) int[][] res = new int[K][2]; if(points == null || points.length == 0 || K == 0) return res; PriorityQueue<int[]> maxHeap = new PriorityQueue<>(new Comparator<int[]>() @Override public int compare(int[] p1, int[] p2) double d1 = getDistance(p1, new int[] 0, 0); double d2 = getDistance(p2, new int[] 0, 0); if(d1 == d2) return 0; return d1 < d2 ? 1 : -1; ); for(int[] point : points) if(maxHeap.size() < K) maxHeap.offer(point); else if(getDistance(point, new int[] 0, 0) < getDistance(maxHeap.peek(), new int[] 0, 0)) maxHeap.poll(); maxHeap.offer(point); for(int i = K - 1; i >= 0; i--) res[i] = maxHeap.poll(); return res; public double getDistance(int[] p, int[] o) int x = p[0] - o[0]; int y = p[1] - o[1]; return Math.sqrt(x * x + y * y);
M3: quick select
time = O(n), space = O(logn)
class Solution int[] origin = 0, 0; public int[][] kClosest(int[][] points, int K) int[][] res = new int[K][2]; if(points == null || points.length == 0 || K == 0) return res; quickSelect(points, 0, points.length - 1, K - 1); for(int i = 0; i < K; i++) res[i] = points[i]; return res; public void quickSelect(int[][] arr, int left, int right, int target) int mid = partition(arr, left, right); if(mid == target) return; else if(mid < target) quickSelect(arr, mid + 1, right, target); else quickSelect(arr, left, mid - 1, target); public int partition(int[][] arr, int left, int right) int[] pivot = arr[right]; int start = left, end = right - 1; while(start <= end) if(getDistance(arr[start], origin) < getDistance(pivot, origin)) start++; else if(getDistance(arr[end], origin) >= getDistance(pivot, origin)) end--; else swap(arr, start++, end--); swap(arr, start, right); return start; public void swap(int[][] arr, int i, int j) int[] tmp = arr[i]; arr[i] = arr[j]; arr[j] = tmp; public double getDistance(int[] p, int[] o) int x = p[0] - o[0]; int y = p[1] - o[1]; return Math.sqrt(x * x + y * y);
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