Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

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探讨在二维平面上，给定n个点时，如何找到并计算最多有多少个点位于同一直线上。这是一个经典的几何问题，涉及到点、线等基本元素的数学关系。
  • Given n points on a 2D plane, find the maximum number of points that lie on the same
    straight line.
    在这里插入图片描述
    在这里插入图片描述
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Note:[p0] = CmNote:whereand θsims=zλdwhere:∗Where FT−1where g(ω)Note:where, INote :# OptionalNote:τc≈10−11sτc≈10−11swhere W(w)
Leetcode--->Given n points on a 2D plane, find the maximum number of points that lie on the same str
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Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. /** * Definition for a point. * class Point { * int x; * int y; * Point() {
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.
题目描述 Given n points on a 2D plane, find the maximum number of points that lie on the same straight
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import java.util.*; public class Solution {     public int maxPoints(Point[] points) {         if(points.length             return points.length;         int flag=0;         for(int i=0;i      
LeetCode-Given n points on a 2D plane, find the maximum number of points that lie on the same straig
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Given n points on a 2D plane, find the maximum number of points that lie on the same straight line 题意为：给定二维平面上的n个点，找到位于同一直线上的最大点数。 这道题给了我们一堆二维点，然后让我们求最大的共线点的个数，根据初中数学我们知道，两点确定一条直线，而且可以写成y = ax + b的形...
【LeetCode-Hard-1】【Max Points on a Line】【点集中最大共线数】
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刷题玩。尽量保证2天至少一道题。 题目主要来自LeetCode Hard题目，以及Acceptance小于20%的Medium/Easy题目（目前共80题）。
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n points on a 2D plane, find the maximum number of points that lie on the same straight line(python)
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# Definition for a point. # 寻找最大共线的点的数量 class Point: def __init__(self,a=0.0,b=0.0): self.x = float(a) self.y = float(b) class Solution: # 求取两点间的斜率 def compute_rate(self,p...
leetcode_3:Given n points on a 2D plane, maximum number of points that lie on the same st
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题目： Given n points on a 2D plane, find themaximum number of points that lie on the same straight line. 此题就是要考虑所有的情况： 需要两重循环，第一重循环遍历起始点a，第二重循环遍历剩余点b。 a和b如果不重合，就可以确定一条直线。 对于每个点a，构建 ...
【leetcode每日一题】149.Max Points on a line
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题目：Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 解析：题目要求是给定n个2维的点，求出总共有多少个点在同一条直线上。由数学知识可知，给定三个点a，b，c，如果三个点在一条直线上，则a和b的斜率与c和d的斜率是相同的。用哈希表来做，针对每个点
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Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 任意一条直线都可以表述为 y = ax + b 假设，有两个点(x1,y1), (x2,y2)，如果他们都在这条直线上则有 y1 = kx1 +b y2 = kx2 +b 由此
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Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 法一：两点确定一条直线（两两组合），然后在判断其他点是否在这条直线上，O（n^3） 法二：遍历每一个点i {其他点j与该点i组成直线的斜率xl，斜率相同，统计加1----斜率相同又过同一个点i
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【题目描述】：Given n points on a 2D plane, find the maximum number of points that lie on the same straight line 在一个二维平面上有n个点，求出现在同一直线上的点的最大个数 在此之间，我还去学了一下HashMap 是Map接口的常用实类 创建一个HashMap:
Given 3 points on a 2D plane, design a function to do the following: determine whether 3 points lie on a straight line; if they do not lie on a straight line, draw a triangle with those points. (please import matplotlib) Note that the straight line or one of the triangle's edges may be perpendicular to the x-axis. Please take this possibility into account. The following shows the execution of your program.
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Here is a Python function that takes in 3 points and determines whether they lie on a straight line or not. If they do not lie on a straight line, it uses Matplotlib to draw a triangle with those points: ```python import matplotlib.pyplot as plt def plot_triangle(p1, p2, p3): # Check if the points lie on a straight line if (p2[1]-p1[1])*(p3[0]-p2[0]) == (p3[1]-p2[1])*(p2[0]-p1[0]): print('The points lie on a straight line.') else: # Create a figure and axis object fig, ax = plt.subplots() # Plot the points ax.plot(p1[0], p1[1], 'bo') ax.plot(p2[0], p2[1], 'bo') ax.plot(p3[0], p3[1], 'bo') # Plot the triangle ax.plot([p1[0], p2[0]], [p1[1], p2[1]], 'r-') ax.plot([p2[0], p3[0]], [p2[1], p3[1]], 'r-') ax.plot([p3[0], p1[0]], [p3[1], p1[1]], 'r-') # Set the x and y limits ax.set_xlim(min(p1[0], p2[0], p3[0])-1, max(p1[0], p2[0], p3[0])+1) ax.set_ylim(min(p1[1], p2[1], p3[1])-1, max(p1[1], p2[1], p3[1])+1) # Show the plot plt.show() ``` To use this function, simply call it with the 3 points as input: ```python p1 = (1, 1) p2 = (2, 3) p3 = (4, 2) plot_triangle(p1, p2, p3) ``` This will plot a triangle with the points `(1, 1)`, `(2, 3)`, and `(4, 2)`. If the points lie on a straight line, the function will print "The points lie on a straight line." instead of plotting the triangle.
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