LeetCode Number of Boomerangs 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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给定一堆点集，问回飞镖的个数。一个回飞镖是一个三点集(i,j,k)，且满足i到j和k的距离相等。 如果和i距离相等的点有n个，则这样的三点集有$$A_n^2=n(n-1)$$，也就是从n个点中拿两个点做排列。所以问题就转换为对每个点，都求其他所有点和该点的距离，求到距离相等的点的个数n，然后代入公式计算。 代码如下： [cpp] class Solution { public: int numberOfBoomerangs(vector<pair<int, int>>& points) { int ans = 0; for (int i = 0; i < points.size(); ++i) { unordered_map<int, int> dist_cnt; for (int j = 0; j < points.size(); ++j) { int xdiff = points[i].first – points[j].first; int ydiff = points[i].second – points[j].second; ++dist_cnt[xdiff*xdiff + ydiff*ydiff]; } for (auto it = dist_cnt.begin(); it != dist_cnt.end(); ++it)ans += it->second*(it->second – 1); } return ans; } }; [/cpp] 本代码提交AC，用时369MS。 因为题中说到所有点都是不相同的，所以第7行没必要限制j!=i，即使j==i，算出来的距离是0，dist_cnt[0]=1，也就是只有点x一个点和自己的距离是0。到第12行计算排列数时，n(n-1)=0，所以没关系。]]>