Given n points in the plane that are all pairwise distinct, a "boomerang" is a tuple of points (i, j, k) such that the distance between iand j equals the distance between i and k (the order of the tuple matters). Find the number of boomerangs. Y…

Given n points in the plane that are all pairwise distinct, a "boomerang" is a tuple of points (i, j, k) such that the distance between iand j equals the distance between i and k (the order of the tuple matters). Find the number of boomerangs. Y…

Given n points in the plane that are all pairwise distinct, a "boomerang" is a tuple of points (i, j, k) such that the distance between iand j equals the distance between i and k (the order of the tuple matters). Find the number of boomerangs. Y…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 [LeetCode] 题目地址:https://leetcode.com/problems/number-of-boomerangs/ Difficulty: Easy 题目描述 Given n points in the plane that are all pairwise distinct, a "boomerang" is a…

Given n points in the plane that are all pairwise distinct, a "boomerang" is a tuple of points (i, j, k) such that the distance between i and j equals the distance between i and k (the order of the tuple matters). Find the number of boomerangs.…

Given n points in the plane that are all pairwise distinct, a "boomerang" is a tuple of points (i, j, k) such that the distance betweeni and j equals the distance between iand k (the order of the tuple matters). Find the number of boomerangs. Yo…

［抄题］: Given n points in the plane that are all pairwise distinct, a "boomerang" is a tuple of points (i, j, k) such that the distance between i and j equals the distance between i and k (the order of the tuple matters). Find the number of boomer…

题目如下: 解题思路:我首先用来时间复杂度是O(n^3)的解法,会判定为超时:后来尝试O(n^2)的解法,可以被AC.对于任意一个点,我们都可以计算出它与其余点的距离,使用一个字典保存每个距离的点的数量,例如dic[2] = 4,表示与该点距离为2的点有四个,那么这四个点任意选两个点就可以和当前点组成Boomerang,根据排列的原理,一共有4*3种方式.依次类推,进而求出当前点与所有不同距离的点组成的Boomerang的数量,最后求出所有点的Boomerang的和. 代码如下: class S…

给定平面上 n 对不同的点,“回旋镖” 是由点表示的元组 (i, j, k) ,其中 i 和 j 之间的距离和 i 和 k 之间的距离相等(需要考虑元组的顺序).找到所有回旋镖的数量.你可以假设 n 最大为 500,所有点的坐标在闭区间 [-10000, 10000] 中.示例:输入:[[0,0],[1,0],[2,0]]输出:2解释:两个回旋镖为 [[1,0],[0,0],[2,0]] 和 [[1,0],[2,0],[0,0]]详见:https://leetcode.com/problems/…

要求 给出平面上n个点,寻找存在多少点构成的三元组(i j k),使得 i j 两点的距离等于 i k 两点的距离 n 最多为500,所有点坐标范围在[-10000, 10000]之间 示例 [[0,0],[1,0],[2,0]] 2 思路 对于每个点i,遍历其余点到i的距离(时间n2,空间n) 查找表记录其他点到 i 的<距离,频次> 对于每个距离,满足条件的三元组个数为:频次 x (频次-1) 求距离时用平方,防止浮点数误差 最长距离,2000^2+2000^2,32位整型不越界 实现 用…