-
Notifications
You must be signed in to change notification settings - Fork 0
/
973.py
45 lines (33 loc) · 1.38 KB
/
973.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
# 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
class Solution:
def kClosest(self, points: 'List[List[int]]',
K: 'int') -> 'List[List[int]]':
index = sorted(
[(v, v[0] * v[0] + v[1] * v[1]) for i, v in enumerate(points)],
key=lambda x: x[1])
return [i[0] for i in index[:K]]
if __name__ == "__main__":
points = [[3, 3], [5, -1], [-2, 4]]
print(Solution().kClosest(points, 2))
points = [[1, 3], [-2, 2]]
K = 1
print(Solution().kClosest(points, K))