[leetcode] 447. Number of Boomerangs @ python_number of boomerangs python-优快云博客

本文链接：https://blog.youkuaiyun.com/danspace1/article/details/87715544

本文介绍了一种算法，用于计算平面上给定点集中的回旋镖数量。回旋镖定义为三个点的元组，其中两个点到第三个点的距离相等。文章详细解释了算法思路，并提供了具体实现代码。

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原题

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 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]]

解法

遍历points, 对每一个点, 构建一个字典, 字典的键是它与其它点的距离, 值是距离的计数, 计算完距离之后, 我们遍历字典, 求出这个点对应的所有的回旋镖的个数, 用排列组合计算.
Time: O(n**2)
Space: O(n)

代码

class Solution:
    def numberOfBoomerangs(self, points: 'List[List[int]]') -> 'int':
        res = 0
        for x in points:
            count = {}
            for y in points:
                d = (x[0]-y[0])**2 + (x[1]-y[1])**2
                count[d] = count.get(d, 0) + 1
            # calculate all the boomerang
            for d in count:
                res += count[d]*(count[d]-1)
        
        return res