Given an array of points where points[i] = [xi, yi] represents a point on the X-Y plane and an integer k, return the k closest points to the origin (0, 0).
The distance between two points on the X-Y plane is the Euclidean distance (i.e., √(x1 - x2)2 + (y1 - y2)2).
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:
1
2
3
4
5
6
7
|
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:
1
2
3
|
Input: points = [[3,3],[5,-1],[-2,4]], k = 2
Output: [[3,3],[-2,4]]
Explanation: The answer [[-2,4],[3,3]] would also be accepted.
|
Constraints:
1 <= k <= points.length <= 104-104 < xi, yi < 104
解题思路：
最直接的解法就是把每个点的距离计算出来之后，然后根据大小排序取前k。
难点：如何在排序算法中采用最快的方式。
可以考虑使用快速排序方法：
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
|
class Solution(object):
def kClosest(self, points, k):
"""
:type points: List[List[int]]
:type k: int
:rtype: List[List[int]]
"""
def calDist(points):
return [(x,x[0]*x[0]+x[1]*x[1]) for x in points]
def quickSort(lists,i,j):
if i >= j:
return list
pivot = lists[i]
low = i
high = j
while i < j:
while i < j and lists[j][1] >= pivot[1]:
j -= 1
lists[i]=lists[j]
while i < j and lists[i][1] <=pivot[1]:
i += 1
lists[j]=lists[i]
lists[j] = pivot
quickSort(lists,low,i-1)
quickSort(lists,i+1,high)
return lists
_points=calDist(points)
_points=quickSort(_points,0,len(_points)-1)
return [_points[idx][0] for idx in range(k)]
|
但是快速排序性能不稳定，性能取决于pivot的选择。因为这个题有限制条件为topK，所以我们可以进一步考虑使用堆排序，通过维护一个大小为K的堆来遍历数据。
这里为了简化代码，我们调用python自带的heapq模块。由于heapq模块默认的保留K最大，所以为了获得k距离最近的点，我们需要把距离取负值。
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
|
import heapq
class Solution(object):
def kClosest(self, points, k):
"""
:type points: List[List[int]]
:type k: int
:rtype: List[List[int]]
"""
def dist(point):
return (-(point[0]**2+point[1]**2),point)
h=[]
for point in points:
if len(h)<k:
heapq.heappush(h,dist(point))
else:
_new=dist(point)
if h[0][0]<_new[0]:
heapq.heappop(h)
heapq.heappush(h,_new)
return [x[1] for x in h]
|
执行用时：148 ms, 在所有 Python 提交中击败了37.37%的用户
内存消耗：18.1 MB, 在所有 Python 提交中击败了53.54%的用户
使用堆在性能上显著优于使用快排。
