447. Number of Boomerangs 难度：easy

题目：

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 andj 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]]

思路：

若在一组点集{a, b, c, d, ...}中，以点a为一个端点，与dis(a, b)相等的点存在n个（包含点b），因为三元组[a, b, c]与三元组[a, c, b]并不相同，所以实际为排列问题，答案为n(n - 1)。

程序：

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