【leetcode】447. Number of Boomerangs【E】
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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]]
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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
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