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: 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…

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. Example 1: Input: [[1,1],[2,2],[3,3]] Output: 3 Explanation: ^ | | o | o | o +-------------> 0 1 2 3 4 Example 2: Input: [[1,1],[3,2],[5,3],[4,1],[2,3…

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…

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 题解: 思路比较简单,每条直线都可以表示为y=kx+b,所以对于任意三点,如果它们共线,那么它们中任意两点的斜率都相等. 所以就遍历points数组,对其中的每一个元素计算它和位于它后面的数组元素的斜率并保存在一个hashmap中. 这个hashmap的键就是两点构成直线的斜率,值就是和当前元素po…

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 给一个由n个点组成的2D平面,找出最多的同在一条直线上的点的个数. 共线点的条件是斜率一样,corn case:点相同:x坐标相同. Java: public class Solution { public int maxPoints(Point[] points) { int res = 0; f…

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 解题思路: 本题主要需要考虑到斜线的情况,可以分别计算出过points[i]直线最多含几个点,然后算出最大即可,由于计算points[i]的时候,前面的点都计算过了,所以不需要把前面的点考虑进去,所以问题可以转化为过points[i]的直线最大点的个数,解题思路是用一个HashMap储存斜率,遍历p…

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.  求二维平面上n个点中,最多共线的点数.     1.比较直观的方法是,三层循环,以任意两点划线,判断第三个点是否在这条直线上.   比较暴力   2.使用map来记录每个点的最大数目.   /** * Definition for a point. * class Point { * int x;…

//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…

1.问题描述 Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 2.翻译 对于在一个平面上的N个点,找出在同一条直线的最多的点的数目 3.思路分析 我们知道任意的两个点可以构成一条直线,对于一条在直线的上的点,他们必然具有相同的斜率,这个我初中都知道了.因为找出最多的点的方法也就比较简单了,我们只要依次遍历这些点,并且记录相同的斜率的点的数目,这样…

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. /** * Definition for a point. * struct Point { * int x; * int y; * Point() : x(0), y(0) {} * Point(int a, int b) : x(a), y(b) {} * }; */ class Solutio…

题目如下: A move consists of taking a point (x, y) and transforming it to either (x, x+y) or (x+y, y). Given a starting point (sx, sy) and a target point (tx, ty), return True if and only if a sequence of moves exists to transform the point (sx, sy) to (…

国庆中秋长假过完,又要开始上班啦.先刷个题目找找工作状态. Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water. Find…

[2]Add Two Numbers (2018年12月23日,review) 链表的高精度加法. 题解:链表专题:https://www.cnblogs.com/zhangwanying/p/9797184.html [7]Reverse Integer (2018年12月23日, review) 给了一个32位的整数,返回它的reverse后的整数.如果reverse后的数超过了整数的范围,就返回 0. Example 1: Input: 123 Output: 321 Example 2:…

[1]Two Sum (2018年11月9日,k-sum专题,算法群衍生题) 给了一个数组 nums, 和一个 target 数字,要求返回一个下标的 pair, 使得这两个元素相加等于 target . 题解:我这次最大范围的优化代码, hash-table + one pass,时间复杂度 O(N),空间复杂度 O(N).重点在于动态找,一边生成hash-table的同时去找答案,不是先生成hash-table再开始找答案. //这种类似 hash 的能用 unordered_map 就不要…

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…

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].…

[LeetCode]554. Brick Wall 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.me/ 题目地址:https://leetcode.com/problems/brick-wall/description/ 题目描述: There is a brick wall in front of you. The wall is rectangular and has…

53. Maximum Subarray[leetcode] Find the contiguous subarray within an array (containing at least one number) which has the largest sum. For example, given the array [-2,1,-3,4,-1,2,1,-5,4],the contiguous subarray [4,-1,2,1] has the largest sum = 6. 挑…

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…

写在前面   老粉丝可能知道现阶段的LeetCode刷题将按照某一个特定的专题进行,之前的[贪心算法]已经结束,虽然只有三个题却包含了简单,中等,困难这三个维度,今天介绍的是第二个专题[数组] 数组(Array)是一种线性表数据结构.它用一组连续的内存空间,来存储一组具有相同类型的数据.在每一种编程语言中,基本都会有数组这种数据类型.不过,它不仅仅是一种编程语言中的数据类型,还是一种最基础的数据结构. 贪心算法回顾: [LeetCode]贪心算法--买卖股票的最佳时机II(122) [LeetC…

Best Time to Buy and Sell Stock with Transaction Fee 题目等级:Medium 题目描述: Your are given an array of integers prices, for which the i-th element is the price of a given stock on day i; and a non-negative integer fee representing a transaction fee. You m…

题目如下: Given an array of scores that are non-negative integers. Player 1 picks one of the numbers from either end of the array followed by the player 2 and then player 1 and so on. Each time a player picks a number, that number will not be available f…

[3]Longest Substring Without Repeating Characters [11]Container With Most Water [15]3Sum (2019年2月26日) 给了一个乱序的数组,返回一个结果数组,数组里面每个元素是一个三元组, 三元组的和加起来为0. 题解:先固定第一个数,然后后面两个数的控制用夹逼定理,2 pointers 来解. class Solution { public: vector<vector<int>> threeSu…

链接:https://leetcode.com/tag/design/ [146]LRU Cache [155]Min Stack [170]Two Sum III - Data structure design (2018年12月6日,周四) Design and implement a TwoSum class. It should support the following operations: add and find. add - Add the number to an inter…

题目如下: Given an array A of positive integers, A[i]represents the value of the i-th sightseeing spot, and two sightseeing spots i and j have distance j - i between them. The score of a pair (i < j) of sightseeing spots is (A[i] + A[j] + i - j) : the su…

题目如下: 解题思路:本题和[leetcode]97. Interleaving String非常相似,同样可以采用动态规划的方法.记dp[i][j] = 1或者0 表示pattern[0:i]是否匹配string[0:j] ,如果pattern[i] == string[j] 或者 pattern[i] == '?',那么dp[i][j]  = dp[i-1][j-1]:如果pattern[i] = '*' 则复杂一些,因为'*'可以匹配s[j-1],s[j] 或者不匹配,因此只要满足其中任意…

题目如下: We are given the root node of a maximum tree: a tree where every node has a value greater than any other value in its subtree. Just as in the previous problem, the given tree was constructed from an list A (root = Construct(A)) recursively with…

题目如下: Given a set of points in the xy-plane, determine the minimum area of any rectangle formed from these points, with sides not necessarily parallel to the x and y axes. If there isn't any rectangle, return 0. Example 1: Input: [[1,2],[2,1],[1,0],[…