[leetcode: Python]447.Number of Boomerangs

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

题目大意：
给定平面上的n个两两不同的点，一个“回飞镖”是指一组点(i, j, k)满足i到j的距离=i到k的距离（考虑顺序）

计算回飞镖的个数。你可以假设n最多是500，并且点坐标范围在 [-10000, 10000] 之内。

解题思路：
枚举点i(x1, y1)，计算点i到各点j(x2, y2)的距离，并分类计数

利用排列组合知识，从每一类距离中挑选2个点的排列数 A(n, 2) = n * (n - 1)

将上述结果累加即为最终答案。

方法一：1135ms

class Solution(object):    def numberOfBoomerangs(self, points):        """        :type points: List[List[int]]        :rtype: int        """        ans = 0        for x1, y1 in points:            dmap = collections.defaultdict(int)            for x2, y2 in points:                dmap[(x1 - x2) ** 2 + (y1 - y2) ** 2] += 1            for d in dmap:                ans += dmap[d] * (dmap[d] - 1)        return ans