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 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]]

 

这道题定义了一种类似回旋镖形状的三元组结构，要求第一个点和第二个点之间的距离跟第一个点和第三个点之间的距离相等。

现在给了n个点，找出回旋镖的个数。如果有n个点和a距离相等，那么排列方式为n(n-1)，也就是可以遍历所有点，让每个点都做一次点a，然后遍历其他所有点，统计和a距离相等的点有多少个，然后分别带入n(n-1)计算结果并累加到res中，只有当n大于等于2时，res值才会真正增加，参见代码如下：

class Solution {
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points) {
        int res = 0;
        for (int i = 0; i < points.size(); ++i) {
            unordered_map<int, int> m;
            for (int j = 0; j < points.size(); ++j) {
                int a = points[i].first - points[j].first;
                int b = points[i].second - points[j].second;
                ++m[a * a + b * b];
            }
            for (auto it = m.begin(); it != m.end(); ++it) {
                res += it->second * (it->second - 1);
            }
        }
        return res;
    }
};