[leetcode] 973. K Closest Points to Origin @ python

原题

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

解法1

构建字典d, 储存距离和点的键值对. 然后将距离转化为list, 使用小根堆弹出最小距离, 将最小距离对应的点加到结果里, 取完K个点的时候就返回结果.

代码

class Solution(object):
    def kClosest(self, points, K):
        """
        :type points: List[List[int]]
        :type K: int
        :rtype: List[List[int]]
        """
        d = collections.defaultdict(list)
        for p in points:
            distance = p[0]**2 + p[1]**2
            d[distance].append(p)
        l = list(d.keys())
        heapq.heapify(l)
        
        count = 0
        ans = []
        while True:
            distance = heapq.heappop(l)
            count += len(d[distance])
            ans.extend(d[distance])
            if count >= K:
                return ans

解法2

参考: https://leetcode.com/problems/k-closest-points-to-origin/discuss/217999/JavaC%2B%2BPython-O(NlogK)
直接使用heapq.nsmallest()方法.

代码

class Solution(object):
    def kClosest(self, points, K):
        """
        :type points: List[List[int]]
        :type K: int
        :rtype: List[List[int]]
        """
        return heapq.nsmallest(K, points, key = lambda (x, y):x*x+y*y)