问题描述:

 

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 jequals the distance between i and k (the order of the tuple matters).

 

Find the number of boomerangs. You may assume that n will be at most 500 and coordinates of points are all in the range [-10000, 10000](inclusive).

 

Example:

 

Input:
[[0,0],[1,0],[2,0]]

Output:
2

Explanation:
The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]]

 

思路:主要学习np.unique()-去重并排序,同时根据return_counts参数可以确定各个元素的个数,squareform()-将点之间距离在简洁和冗余模式相互转化,pdist()-求点之间距离,等函数的应用

代码:

 

 1 import numpy as np
 2 from numpy import *
 3 from scipy.spatial.distance import pdist, squareform
 4 
 5 class Solution:   
 6     def numberOfBoomerangs(self, points: List[List[int]]) -> int:
 7         a = squareform(pdist(np.array(points)))
 8       
 9         
10         result = 0
11         for i in a:#遍历每一行
12             count = np.unique(i,return_counts=True)[1]
13             result += sum(count*(count - 1))
14           
15                     
16         return result

 

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