intLeetCode::maxComDivisor(inta,intb){if(b)returnmaxComDivisor(b,a%b);elsereturna; }intLeetCode::maxPoints(vector<Point>&points){if(points.size() <3)returnpoints.size();intmax =2; auto it=points.cbegin();while(it !=points.cend()){//cur统计it的每个点的对应的直线上的最大点数量;...
Givennpoints on a 2D plane, find the maximum number of points that lie on the same straight line. 思路: (1)点1到点n (a)以点1为参考点,求“点1”与“点2到点n”各个斜率下点的个数; (求出直线最多的点数值max1) (b)以点2为参考点,求“点2”与“点3到点n”各个斜率下点的个数; (求...
* Point(int a, int b) { x = a; y = b; } * } */ public class Solution { public int maxPoints(Point[] points) { if (points.length <= 1) return points.length; int maxUniv = Integer.MIN_VALUE; for (int i = 0; i < points.length; i++) { Point cur = points[i]; Hash...
DescriptionGiven n points on a 2D plane, find the maximum number of points that lie on the same straight line. Example 1: Input: [[1,1],[2,2],[3,3]] Output: 3 Explanation: ^ | | o | o | o +---…
Can you solve this real interview question? Max Points on a Line - Given an array of points where points[i] = [xi, yi] represents a point on the X-Y plane, return the maximum number of points that lie on the same straight line. Example 1: [https://a
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 思路: 对所有可以配对的点计算他们之间的斜率,并保存。时间复杂度O(n^2),空间复杂度为O(n)。 算法: public int maxPoints(Point[] points) { ...
如何在Python中实现计算直线上最多点数的算法? Leetcode上的Max Points on a Line问题有哪些解题思路? 在解决Max Points on a Line问题时,如何处理重复的点? 题目大意 在一个平面上有n个点,求一条直线最多能够经过多少个这些点。 解题思路 哈希表保存斜率 代码 注意有个坑: 测试集[[0,0],[94911151,9491115...
Max Points on a Line 题目本身不难,一次AC可能有点困难,因为要考虑的东西还是挺多的。两层循环,外层遍历所以点,内层遍历外层点之后的所有点,同时在内层循环用一个HashMap来保存每个斜率对应的,这样在内存循环中,斜率相同就代表是在同一条直线上了。这里要注意的有两点: ...
https://leetcode-cn.com/problems/max-points-on-a-line/ Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 题意 给定一个二维平面,平面上有 n 个点,求最多有多少个点在同一条直线上。
https://leetcode.com/problems/max-points-on-a-line/ 思路很简单,就是找出两两点的pair,看有多少pair的slope是一样的。 这里设想plane中有一条line就是我们要求的最多点的line,那么只要我们循环到这条line上的任意一个点,那么就可以找到与其共线的其他所有点,所以每次做二维循环,都reset一个dict,然后求这个...