k-closest-points-to-origin

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Algorithms

Given an array of <code>points</code> where <code>points[i] = [xi, yi]</code> represents a point on the X-Y plane and an integer <code>k</code>, return the <code>k</code> closest points to the origin <code>(0, 0)</code>.

The distance between two points on the X-Y plane is the Euclidean distance (i.e., <code>&radic;(x1 - x2)2 + (y1 - y2)2</code>).

You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in).

&nbsp;
Example 1:
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/03/closestplane1.jpg" style="width: 400px; height: 400px;" />
<pre>
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) &lt; 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]].
</pre>

Example 2:

<pre>
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.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= k &lt;= points.length &lt;= 104</code></li>
	<li><code>-104 &lt;= xi, yi &lt;= 104</code></li>
</ul>

kClosest

K Closest Points to Origin

Kth Largest Element in an Array

kth-largest-element-in-an-array

数组中的第K个最大元素

Top K Frequent Elements

top-k-frequent-elements

前 K 个高频元素

Top K Frequent Words

top-k-frequent-words

前K个高频单词

Find Nearest Point That Has the Same X or Y Coordinate

find-nearest-point-that-has-the-same-x-or-y-coordinate

找到最近的有相同 X 或 Y 坐标的点

Minimum Rectangles to Cover Points

minimum-rectangles-to-cover-points

覆盖所有点的最少矩形数目

K-th Nearest Obstacle Queries

k-th-nearest-obstacle-queries

第 K 近障碍物查询

Geometry

Array

Math

Divide and Conquer

Quickselect

Sorting

Heap (Priority Queue)

给定一个数组 <code>points</code>&nbsp;，其中&nbsp;<code>points[i] = [xi, yi]</code>&nbsp;表示 X-Y 平面上的一个点，并且是一个整数 <code>k</code> ，返回离原点 <code>(0,0)</code> 最近的 <code>k</code> 个点。

这里，平面上两点之间的距离是&nbsp;欧几里德距离（&nbsp;<code>√(x1&nbsp;- x2)2&nbsp;+ (y1&nbsp;- y2)2</code>&nbsp;）。

你可以按 任何顺序 返回答案。除了点坐标的顺序之外，答案 确保 是 唯一 的。

&nbsp;

示例 1：

<img src="https://assets.leetcode.com/uploads/2021/03/03/closestplane1.jpg" style="height: 400px; width: 400px;" />

<pre>
输入：points = [[1,3],[-2,2]], k = 1
输出：[[-2,2]]
解释： 
(1, 3) 和原点之间的距离为 sqrt(10)，
(-2, 2) 和原点之间的距离为 sqrt(8)，
由于 sqrt(8) &lt; sqrt(10)，(-2, 2) 离原点更近。
我们只需要距离原点最近的 K = 1 个点，所以答案就是 [[-2,2]]。
</pre>

示例 2：

<pre>
输入：points = [[3,3],[5,-1],[-2,4]], k = 2
输出：[[3,3],[-2,4]]
（答案 [[-2,4],[3,3]] 也会被接受。）
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= k &lt;= points.length &lt;= 104</code></li>
	<li><code>-104&nbsp;&lt; xi, yi&nbsp;&lt; 104</code></li>
</ul>