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

题目地址：K Closest Points to Origin - LeetCode

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

这道题目是找前TopK小的元素，最容易想到的方法是全排序，然后取前K个。
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 resfor 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

简化一下：

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);}
};

LeetCode 973. K Closest Points to Origin--TopK 问题--最小堆--C++,Python解法相关推荐

Leetcode PHP题解--D29 973. K Closest Points to Origin
973. K Closest Points to Origin 题目链接 973. K Closest Points to Origin 题目分析给一个坐标数组points,从中返回K个离0,0最近 ...
LeetCode Daily challenge - K Closest Points to Origin
文章目录题目大意思路代码总结题目大意找前K小元素,只不过元素是二维的坐标思路很容易想到排序后取前K个,虽然可以但是速度不够快.实际上类似于找第K大(小),可以用划分的方法找,每次随机划 ...
merge k sorted lists java_LeetCode 第23题 Merge k Sorted Lists【分而治之】【最小堆】（Java）...
这道题的题目是合并k个有序的链表,标定难度为Hard,详细需求: 合并k个有序的链表,返回一个新的有序链表. 例子: Input: 1->4->5, 1->3->4, 2-&g ...
LeetCode 159. Longest Substring with At Most Two Distinct Characters --Java,C++,Python解法
题目地址:Longest Substring with At Most Two Distinct Characters - LeetCode Given a string s , find the l ...
LeetCode 673. Number of Longest Increasing Subsequence--O(N log N )--Java,C++,Python解法
题目地址:Number of Longest Increasing Subsequence - LeetCode 做这道题目前建议先做:Longest Increasing Subsequence - ...
LeetCode 142. Linked List Cycle II--单向链表成环的起点--C++,Python解法
题目地址:Linked List Cycle II - LeetCode Given a linked list, return the node where the cycle begins. If ...
LeetCode题解(1647)：字符频次唯一的最小删除次数(Python)
题目:原题链接(中等) 标签:贪心算法.排序解法时间复杂度空间复杂度执行用时 Ans 1 (Python) O(NlogN)O(NlogN)O(NlogN) O(N)O(N)O(N) 132m ...
LeetCode刷题之python解法（持续更新）
1. Two Sum 4行 class Solution:def twoSum(self, nums: List[int], target: int) -> List[int]:d = {}fo ...
LeetCode 973. 最接近原点的 K 个点（排序/优先队列/快排）
文章目录 1. 题目 2. 解题 2.1 排序 2.2 优先队列 2.3 快排思路 1. 题目我们有一个由平面上的点组成的列表 points.需要从中找出 K 个距离原点 (0, 0) 最近的点. ...

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

LeetCode 973. K Closest Points to Origin--TopK 问题--最小堆--C++,Python解法相关推荐

最新文章

热门文章