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‎README.adoc

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//|{counter:codes}
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//|{leetcode_base_url}/max-points-on-a-line/[149. Max Points on a Line^]
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//|{source_base_url}/_0149_MaxPointsOnALine.java[Java]
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//|{doc_base_url}/0149-max-points-on-a-line.adoc[题解]
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//|Hard
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//|
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|{counter:codes}
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|{leetcode_base_url}/max-points-on-a-line/[149. Max Points on a Line^]
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|{source_base_url}/_0149_MaxPointsOnALine.java[Java]
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|{doc_base_url}/0149-max-points-on-a-line.adoc[题解]
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|{counter:codes}
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|{leetcode_base_url}/evaluate-reverse-polish-notation/[150. Evaluate Reverse Polish Notation^]

‎docs/0149-max-points-on-a-line.adoc

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[#0149-max-points-on-a-line]
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= 149. Max Points on a Line
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= 149. 直线上最多的点数
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{leetcode}/problems/max-points-on-a-line/[LeetCode - Max Points on a Line^]
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Given _n_ points on a 2D plane, find the maximum number of points that lie on the same straight line.
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https://leetcode.cn/problems/max-points-on-a-line/[LeetCode - 149. 直线上最多的点数 ^]
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*Example 1:*
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给你一个数组 `points` ，其中 `points[i] = [x~i~, y~i~]` 表示 *X-Y* 平面上的一个点。求最多有多少个点在同一条直线上。
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[subs="verbatim,quotes,macros"]
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----
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*Input:* [[1,1],[2,2],[3,3]]
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*Output:* 3
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*Explanation:*
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^
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|
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| o
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| o
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| o
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+------------->
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0 1 2 3 4
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----
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*示例 1：*
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*Example 2:*
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image::images/0149-01.jpg[{image_attr}]
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[subs="verbatim,quotes,macros"]
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----
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*Input:* [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
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*Output:* 4
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*Explanation:*
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^
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|
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| o
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| o o
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| o
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| o o
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+------------------->
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0 1 2 3 4 5 6
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----
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....
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输入：points = [[1,1],[2,2],[3,3]]
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输出：3
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....
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*示例 2：*
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image::images/0149-02.jpg[{image_attr}]
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....
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输入：points = [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
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输出：4
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....
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*提示：*
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* `+1 <= points.length <= 300+`
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* `points[i].length == 2`
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* `-10^4^ \<= x~i~, y~i~ \<= 10^4^`
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* `points` 中的所有点 *互不相同*
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*NOTE:* input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
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== 思路分析
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先确定两个点，然后再选一个点，计算三个点的斜率来查看是否在同一条直线。
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[stem]
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++++
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(y_1 - y_2) / (x_1 - x_2) = (y_2 - y_3)/(x_2 - x_3)
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++++
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等价于
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[stem]
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++++
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(x_2 - x_3)* (y_1 - y_2) = (x_1 - x_2) * (y_2 - y_3)
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++++
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这样就避免计算除法。
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[[src-0149]]
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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}/_0149_MaxPointsOnALine.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}/_0149_MaxPointsOnALine_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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‎docs/images/0149-01.jpg

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‎docs/images/0149-02.jpg

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‎docs/index.adoc

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// include::0149-max-points-on-a-line.adoc[leveloffset=+1]
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include::0149-max-points-on-a-line.adoc[leveloffset=+1]
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‎logbook/202503.adoc

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|{counter:codes2503}
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|{leetcode_base_url}/candy/[135. 分发糖果^]
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|{doc_base_url}/0135-candy.adoc[题解]
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|⭕️初步设想是一次遍历。后来看题解，可以从两端遍历，再就和。这样可以省去很多判断。
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|⭕️ 初步设想是一次遍历。后来看题解，可以从两端遍历，再就和。这样可以省去很多判断。
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|{counter:codes2503}
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|{leetcode_base_url}/max-points-on-a-line/[149. 直线上最多的点数^]
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|{doc_base_url}/0149-max-points-on-a-line.adoc[题解]
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|⭕️ 通过计算 stem:[(y_1 - y_2) / (x_1 - x_2) = (y_2 - y_3)/(x_2 - x_3)] 即可确定三个点是否在同一个直线上。
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‎src/main/java/com/diguage/algo/leetcode/_0149_MaxPointsOnALine.java

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package com.diguage.algo.leetcode;
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public class _0149_MaxPointsOnALine {
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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 2025-06-07 21:55:30
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*/
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public int maxPoints(int[][] points) {
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int length = points.length;
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int result = 1;
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for (int i = 0; i < length; i++) {
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int[] a = points[i];
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for (int j = i + 1; j < length; j++) {
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int[] b = points[j];
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int cnt = 2;
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for (int k = j + 1; k < length; k++) {
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int[] c = points[k];
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int m1 = (b[0] - c[0]) * (a[1] - b[1]);
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int m2 = (a[0] - b[0]) * (b[1] - c[1]);
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if (m1 == m2) {
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cnt++;
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}
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}
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result = Math.max(result, cnt);
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}
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}
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return result;
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}
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// end::answer[]
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public static void main(String[] args) {
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new _0149_MaxPointsOnALine()
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.maxPoints(new int[][]{{3, 3}, {1, 4}, {1, 1}, {2, 1}, {2, 2}});
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}
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}

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