QUESTIONGiven n points on a 2D plane, find the maximum number of points that lie on the same straight line. 1ST TRY /** * Definition for a point. * struct Point { * int x; * int y; * Point() : x(0), y(0) {} * Point(int a, int b) : x(a), y(b) {} * };…
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…
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. 示例 1: 输入: [[1,1],[2,2],[3,3]] 输出: 3 解释: ^ | | o | o | o +-------------> 0 1 2 3 4 示例 2: 输入: [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]] 输出: 4 解释: ^ | | o | o o | o | o o +------------------…
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…
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. SOLUTION 1: 全部的点扫一次,然后计算每一个点与其它点之间的斜率. 创建一个MAP, KEY-VALUE是 斜率:线上的点的数目. 另外,注意重合的点,每次都要累加到每一条线上. 注意: 1. k = 0 + (double)(points[i].…
Max Points on a Line Submission Details 27 / 27 test cases passed. Status: Accepted Runtime: 472 ms Submitted: 0 minutes ago Submitted Code Language: java Edit Code /** * Definition for a point. * class Point { * int x; * int y; * Point() {…
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. Have you met this question in a real interview? Example Given 4 points: (1,2), (3,6), (0,0), (1,3). The maximum number is 3. LeetCode上的原题,请参见我之前的博…
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. 点和方向确定一条直线. 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 斜率->点数 的map. (1)b与a重合,以a起始的所有直线点数+1 (用dup统一相加) (2)b与a不重…