题目如下: You are given an array coordinates, coordinates[i] = [x, y], where [x, y] represents the coordinate of a point. Check if these points make a straight line in the XY plane. Example 1: Input: coordinates = [[1,2],[2,3],[3,4],[4,5],[5,6],[6,7]] Ou…

2019-10-21 10:35:33 问题描述: 问题求解: public boolean checkStraightLine(int[][] coordinates) { int n = coordinates.length; if (n == 2) return true; for (int i = 3; i < n; i++) { if (area(coordinates[0], coordinates[1], coordinates[i]) != 0) return false; }…

//定义二维平面上的点struct Point { int x; int y; Point(, ):x(a),y(b){} }; bool operator==(const Point& left, const Point& right) { return (left.x==right.x && left.y==right.y); } //求两个点连接成的直线所对应的斜率 double line_equation(const Point& P1, const Poi…

Problem: Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. Suppose that the structure Point is already defined in as following: /** * Definition for a point. * struct Point { * int x; * int y; * Point…

转自:http://peter.sh/experiments/chromium-command-line-switches/ There are lots of command lines which can be used with the Google Chrome browser. Some change behavior of features, others are for debugging or experimenting. This page lists the availabl…

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 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的形式,所有共线的点都满足这个公式.所以这些给定点两两之间都可以算一个斜率,每个斜率代表一条直线,对每一条直线,带入所有的点看是否共线并计算个数,这是整体的思路.但是还有…

题目链接 Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 分析:首先要注意的是,输入数组中可能有重复的点.由于两点确定一条直线,一个很直观的解法是计算每两个点形成的直线,然后把相同的直线合并,最后包含点最多的直线上点的个数就是本题的解.我们知道表示一条直线可以用斜率和y截距两个浮点数(垂直于x轴的直线斜率为无穷大,截距用x截距),同时还需要保存每…

Max Points on a Line 题目描述: Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 解题思路: 1.首先由这么一个O(n^3)的方法,也就是算出每条线的方程(n^2),然后判断有多少点在每条线上(N).这个方法肯定是可行的,只是复杂度太高2.然后想到一个O(N)的,对每一个点,分别计算这个点和其他所有点构成的斜率,具有相同斜率最…

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 思路 关键是浮点数做key不靠谱,struct hash以及 int calcGCD(int a, int b)的写法 代码 /** * Definition for a point. * struct Point { * int x; * int y; * Point() : x(0), y(0)…

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 思路: 自己脑子当机了,总是想着斜率和截距都要相同.但实际上三个点是一条直线的话只要它们的斜率相同就可以了,因为用了相同的参照点,截距一定是相同的. 大神的做法: 对每一个点a, 找出所有其他点跟a的连线斜率,相同为同一条线,记录下通过a的点的线上最大的点数. 找出每一个点的最大连线通过的点数. 其…