題目

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]]

難度

Easy

方法

對(duì)于每個(gè)point，用一個(gè)map統(tǒng)計(jì)各個(gè)距離d下對(duì)應(yīng)的其他point的個(gè)數(shù)n，即map的key為距離d，value為距離該point為d的其他point的個(gè)數(shù)n。然后An取2，即n*(n-1)。最后將各個(gè)point對(duì)應(yīng)的n*(n-1)累加即可

python代碼

class Solution(object):
    def numberOfBoomerangs(self, points):
        """
        :type points: List[List[int]]
        :rtype: int
        """
        result = 0
        for point_i in points:
            distance_map = {}
            for point_j in points:
                distance = (point_i[0] - point_j[0]) * (point_i[0] - point_j[0]) + \
                           (point_i[1] - point_j[1]) * (point_i[1] - point_j[1])
                distance_map[distance] = distance_map.get(distance, 0) + 1

            for distance in distance_map:
                count = distance_map[distance]
                result += count * (count - 1)

        return result

assert Solution().numberOfBoomerangs([[0,0], [1,0], [2,0]]) == 2