题目如下：

We have a list of points on the plane.  Find the Kclosest 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 = 1Output: [[-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 = 2Output: [[3,3],[-2,4]](The answer [[-2,4],[3,3]] would also be accepted.)

 

Note:

1. 1 <= K <= points.length <= 10000
2. -10000 < points[i][0] < 10000
3. -10000 < points[i][1] < 10000

解题思路：太简单了，没啥说的。

代码如下：

class Solution(object):    def kClosest(self, points, K):        """        :type points: List[List[int]]        :type K: int        :rtype: List[List[int]]        """        l = []        for x,y in points:            l.append((x,y,x*x+y*y))        def cmpf(v1,v2):            return v1[2] - v2[2]        l = sorted(l,cmp=cmpf)[:K]        res = []        for x,y,z in l:            res.append([x,y])        return res