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.)
Solution
这个题的思路就是计算每个点的平方和，然后按从小到大排列，取出前个数。我们用优先队列来实现。
Python中的heapq
import heapq
class Solution:
def kClosest(self, points: List[List[int]], K: int) -> List[List[int]]:
q=[]
for x, y in points:
t = (-x*x-y*y,x,y)
if len(q)<K:
heapq.heappush(q,t)
else:
heapq.heappushpop(q,t)
res =[]
for _,x,y in q:
res.append([x,y])
return res
