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‎README.adoc‎

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//|{doc_base_url}/0972-equal-rational-numbers.adoc[题解]
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//|Hard
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//|
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//
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//|{counter:codes}
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//|{leetcode_base_url}/k-closest-points-to-origin/[973. K Closest Points to Origin^]
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//|{source_base_url}/_0973_KClosestPointsToOrigin.java[Java]
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//|{doc_base_url}/0973-k-closest-points-to-origin.adoc[题解]
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//|Medium
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//|
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//
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|{counter:codes}
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|{leetcode_base_url}/k-closest-points-to-origin/[973. K Closest Points to Origin^]
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|{source_base_url}/_0973_KClosestPointsToOrigin.java[Java]
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|{doc_base_url}/0973-k-closest-points-to-origin.adoc[题解]
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|Medium
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|
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//|{counter:codes}
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//|{leetcode_base_url}/subarray-sums-divisible-by-k/[974. Subarray Sums Divisible by K^]
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//|{source_base_url}/_0974_SubarraySumsDivisibleByK.java[Java]

‎docs/0973-k-closest-points-to-origin.adoc‎

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[#0973-k-closest-points-to-origin]
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= 973. K Closest Points to Origin
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= 973. 最接近原点的 K 个点
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{leetcode}/problems/k-closest-points-to-origin/[LeetCode - K Closest Points to Origin^]
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https://leetcode.cn/problems/k-closest-points-to-origin/[LeetCode - 973. 最接近原点的 K 个点^]
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We have a list of `points` on the plane. Find the `K` closest points to the origin `(0, 0)`.
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给定一个数组 `points`，其中 `points[i] = [x~i~, y~i~]` 表示 *X-Y* 平面上的一个点，并且是一个整数 `k`，返回离原点 `(0, 0)` 最近的 `k` 个点。
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(Here, the distance between two points on a plane is the Euclidean distance.)
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这里，平面上两点之间的距离是 *欧几里德距离* stem:[\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}]。
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You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in.)
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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。
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*示例 1：*
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image::images/0973-01.jpg[{image_attr}]
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*Example 1:*
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[subs="verbatim,quotes,macros"]
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----
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*Input:* points = [[1,3],[-2,2]], K = 1
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*Output:* [[-2,2]]
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*Explanation:*
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The distance between (1, 3) and the origin is sqrt(10).
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The distance between (-2, 2) and the origin is sqrt(8).
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Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin.
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We only want the closest K = 1 points from the origin, so the answer is just [[-2,2]].
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----
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....
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输入：points = [[1,3],[-2,2]], k = 1
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输出：[[-2,2]]
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解释：
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(1, 3) 和原点之间的距离为 sqrt(10)，
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(-2, 2) 和原点之间的距离为 sqrt(8)，
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由于 sqrt(8) < sqrt(10)，(-2, 2) 离原点更近。
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我们只需要距离原点最近的 K = 1 个点，所以答案就是 [[-2,2]]。
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....
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*示例 2：*
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*Example 2:*
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[subs="verbatim,quotes,macros"]
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----
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*Input:* points = [[3,3],[5,-1],[-2,4]], K = 2
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*Output:* [[3,3],[-2,4]]
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(The answer [[-2,4],[3,3]] would also be accepted.)
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----
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....
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输入：points = [[3,3],[5,-1],[-2,4]], k = 2
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输出：[[3,3],[-2,4]]
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（答案 [[-2,4],[3,3]] 也会被接受。）
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....
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*提示：*
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*Note:*
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* `1 \<= k \<= points.length \<= 10^4^`
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* `-10^4^ < x~i~, y~i~ < 10^4^`
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. `1 <= K <= points.length <= 10000`
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. `-10000 < points[i][0] < 10000`
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. `-10000 < points[i][1] < 10000`
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== 思路分析
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排序，堆，快速选择等多种解法均可。
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WARNING: 快速选择写的还是漏洞百出！！
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[[src-0973]]
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[tabs]
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====
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一刷::
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+
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--
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[{java_src_attr}]
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----
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include::{sourcedir}/_0973_KClosestPointsToOrigin.java[tag=answer]
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----
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--
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// 二刷::
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// +
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// --
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// [{java_src_attr}]
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// ----
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// include::{sourcedir}/_0973_KClosestPointsToOrigin_2.java[tag=answer]
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// ----
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// --
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====
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== 参考资料
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. https://leetcode.cn/problems/k-closest-points-to-origin/solutions/478516/kuai-lai-miao-dong-topkkuai-pai-bian-xing-da-gen-d/[973. 最接近原点的 K 个点 - 快来秒懂 TopK！快排变形 & 大根堆^]
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. https://leetcode.cn/problems/k-closest-points-to-origin/solutions/477916/zui-jie-jin-yuan-dian-de-k-ge-dian-by-leetcode-sol/[973. 最接近原点的 K 个点 - 官方题解^]
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‎docs/images/0973-01.jpg‎

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‎docs/index.adoc‎

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// include::0973-k-closest-points-to-origin.adoc[leveloffset=+1]
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include::0973-k-closest-points-to-origin.adoc[leveloffset=+1]
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// include::0974-subarray-sums-divisible-by-k.adoc[leveloffset=+1]
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‎logbook/202601.adoc‎

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|{doc_base_url}/0154-find-minimum-in-rotated-sorted-array-ii.adoc[题解]
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|⭕️ 二分查找。重点要看，你关注的点在哪里，如果简化判定。
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|{counter:codes}
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|{leetcode_base_url}/k-closest-points-to-origin/[973. 最接近原点的 K 个点^]
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|{doc_base_url}/0973-k-closest-points-to-origin.adoc[题解]
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|⭕️ 快速选择！快速选择还是写的漏洞百出！
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|===
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‎logbook/corrections.adoc‎

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|{doc_base_url}/0886-possible-bipartition.adoc[题解]
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|❌
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|{counter:codes}
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|{leetcode_base_url}/k-closest-points-to-origin/[973. 最接近原点的 K 个点^]
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|{doc_base_url}/0973-k-closest-points-to-origin.adoc[题解]
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|⭕️
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|===
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截止目前，本轮练习一共完成 {corrections} 道题。

‎src/main/java/com/diguage/algo/leetcode/_0973_KClosestPointsToOrigin.java‎

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package com.diguage.algo.leetcode;
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import java.util.Arrays;
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public class _0973_KClosestPointsToOrigin {
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2026-08-29 21:55:28
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*/
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public int[][] kClosest(int[][] points, int k) {
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quickSelect(points, k - 1, 0, points.length - 1);
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return Arrays.copyOfRange(points, 0, k);
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}
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private void quickSelect(int[][] points, int idx, int left, int right) {
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if (left >= right) {
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return;
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}
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int p = partittion(points, left, right);
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if (p == idx) {
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return;
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}
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if (p < idx) {
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quickSelect(points, idx, p + 1, right);
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} else {
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quickSelect(points, idx, left, p - 1);
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}
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}
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private int partittion(int[][] points, int left, int right) {
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int[] v = points[left];
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int dist = v[0] * v[0] + v[1] * v[1];
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int l = left, r = right + 1;
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while (l <= r) {
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while (++l <= right && points[l][0] * points[l][0] + points[l][1] * points[l][1] < dist) {
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}
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while (left <= --r && points[r][0] * points[r][0] + points[r][1] * points[r][1] > dist) {
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}
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if (l >= r) {
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break;
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}
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swap(points, l, r);
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}
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points[left] = points[r];
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points[r] = v;
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return r;
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}
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private void swap(int[][] points, int a, int b) {
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int[] tmp = points[a];
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points[a] = points[b];
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points[b] = tmp;
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}
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// end::answer[]
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// public int[][] kClosest(int[][] points, int k) {
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// quickSort(points, k, 0, points.length - 1);
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// int[][] result = new int[k][2];
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// System.arraycopy(points, 0, result, 0, k);
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// return result;
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// }
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//
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// private void quickSort(int[][] nums, int k, int left, int right) {
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// if (left >= right) {
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// return;
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// }
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// int l = left, r = right;
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// int pivot = left + ((right - left) / 2);
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// int pn = nums[pivot][0] * nums[pivot][0] + nums[pivot][1] * nums[pivot][1];
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// while (l <= r) {
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// while (l < r
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// && nums[l][0] * nums[l][0] + nums[l][1] * nums[l][1] < pn) {
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// l++;
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// }
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// while (l < r
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// && pn < nums[r][0] * nums[r][0] + nums[r][1] * nums[r][1]) {
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// r--;
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// }
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// if (l <= r) {
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// int[] temp = nums[l];
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// nums[l] = nums[r];
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// nums[r] = temp;
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// l++;
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// r--;
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// }
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//
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// }
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// if (k <= r) {
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// quickSort(nums, k, left, r);
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// } else if (k >= l) {
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// quickSort(nums, k, l, right);
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// }
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// }
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// end::answer[]
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static void main() {
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new _0973_KClosestPointsToOrigin()
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.kClosest(new int[][]{{3, 3}, {5, -1}, {-2, 4}}, 1);
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// .kClosest(new int[][]{{1, 3}, {-2, 2}}, 1);
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}
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}

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