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:

1
2
3
4
5
6
7
8
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

我们求解最简单的欧氏距离，进行n^2次的遍历，将各个点之间的距离存到哈希表当中。如果哈希表的距离统计值超过2，就需要进行计算，得出最终的结果。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
class Solution(object):
    def numberOfBoomerangs(self, points):
        """
        :type points: List[List[int]]
        :rtype: int
        """
        length = len(points)
        res = 0
        for i in range(length):
            distlist = []
            stats = {}
            for j in range(length):
                if (i != j):
                    dist = (points[i][0] - points[j][0]) * (points[i][0] - points[j][0]) + (points[i][1] - points[j][1]) * (points[i][1] - points[j][1])
                    distlist.append(dist)
            # Program optimization: Do not use the 'count' function,
            # Using Hashtable is much better, since the cost for "count" function is o(n).
            for k in distlist:
                if k in stats:
                    stats[k] += 1
                else:
                    stats[k] = 1
            # Count the final number of boomerangs
            # The number forms an arithmetic sequence.
            for m in stats.values():
                if (m == 2):
                    res += m
                elif (m > 2):
                    res += (m * (m-1))
                else:
                    res = res
        return res