标签：loser output min between lan uniq close select double
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); } }
973. K Closest Points to Origin - Medium
标签：loser output min between lan uniq close select double
原文地址：https://www.cnblogs.com/fatttcat/p/11179329.html
