LeetCode 973. K Closest Points to Origin--TopK 問題--最小堆--C++,Python解法

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LeetCode題解專欄：LeetCode題解
我做的所有的LeetCode的題目都放在這個專欄裏，大部分題目Java和Python的解法都有。

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題目地址：K Closest Points to Origin - LeetCode

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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

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這道題目是找前TopK小的元素，最容易想到的方法是全排序，然後取前K個。
Python解法如下：

class Solution:
    def kClosest(self, points: List[List[int]], K: int) -> List[List[int]]:
        res = []
        if len(points) == 0:
            return res
        for i in points:
            i.append(i[0]**2+i[1]**2)
        points.sort(key=lambda x: x[2])
        for i in range(0, K):
            res.append([points[i][0],points[i][1]])
        return res

簡化一下：

class Solution:
    def kClosest(self, points: List[List[int]], K: int) -> List[List[int]]:
        return sorted(points,key = lambda x:x[0]*x[0]+x[1]*x[1])[:K]

使用最小堆可以稍微快一些：

import heapq
class Solution:
    def kClosest(self, points: List[List[int]], K: int) -> List[List[int]]:
        return heapq.nsmallest(K,points,key=lambda x: x[0]*x[0]+x[1]*x[1])

C++解法如下：

class Solution
{
public:
    vector<vector<int>> kClosest(vector<vector<int>> &points, int K)
    {
        sort(points.begin(), points.end(), [](vector<int> &a, vector<int> &b) {
            return a[0] * a[0] + a[1] * a[1] < b[0] * b[0] + b[1] * b[1];
        });
        return vector<vector<int>>(points.begin(), points.begin() + K);
    }
};