【leetcode】447. Number of Boomerangs【E】

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 iand 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]]

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有三个问题

第一个，最开始的时候还去计算了开平方，但实际上并没有用，而且竟然还用了numpy，实在是不开窍

第二个，最开始的时候虽然结果对，但是一直超时，或者超内存，看了一眼网上的结果，发现，其实并不需要在外面弄一个打的dic，每个循环一个dic就行了

第三个，每个循环，每个节点与其他节点比较就行了

class Solution(object):

    def numberOfBoomerangs(self, points):
        p = points

        res = 0
        for i in xrange(len(p)):
            dic = {}
            for j in xrange(len(p)):

                dist = (p[j][1] - p[i][1])**2 +  (p[j][0] - p[i][0])**2
                dic[dist] = dic.get(dist,0) +1

            #print dic
            for i in dic.values():
                if i > 1:
                    res += i*(i-1)


        return res