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number-of-boomerangs.py
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57 lines (41 loc) · 1.6 KB
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'''
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. 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]]
'''
class Solution:
# Approach one
# def recursion(self,n): #'定义递归函数实现求阶乘功能'
# if n==1:
# return 1
# else:
# return n * self.recursion(n-1)
# def numberOfBoomerangs(self, points: List[List[int]]) -> int:
# res = 0
# for i, n in enumerate(points):
# dic = {}
# for j, m in enumerate(points):
# ans = abs(m[0] - n[0])**2 + abs(m[1] - n[1])**2
# dic[ans] = dic.get(ans,0) + 1
# for v in dic.values():
# if v > 1:
# if v >= 4:
# res += self.recursion(v)//2
# else:
# res += self.recursion(v)
# return res
# Approach two
def numberOfBoomerangs(self, points: List[List[int]]) -> int:
res = 0
from collections import Counter
fouc = lambda x, y : x**2 + y**2
for n in points:
for v in Counter([ fouc(m[0] - n[0] , m[1] - n[1]) for m in points]).values():
res += v * (v-1)
return res