[leetcode]: 447. Number of Boomerangs

1.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 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).

boomerangs定义为一个三个点组成的元组[i,j,k]，其中dist[i,j]=dist[i,k]，强调顺序。
给出一组无重复的点，找出其中可以组成boomerangs的对数。
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]]

2.分析

1）因为boomerangs强调顺序，所以每次以一个定点为i，然后求i与其他点之间的距离。
2）求boomerangs的数量。假设求得的距离中有x个点与i点的距离相同，这x个点可以组成的boomerangs数量为

3.代码

class Solution {public:    int distance(pair<int, int> x, pair<int, int> y) {        return (x.first - y.first)*(x.first - y.first) + (x.second - y.second)*(x.second - y.second);    }    int numberOfBoomerangs(vector<pair<int, int>>& points) {        int n = points.size();        int counts = 0;        for (int i = 0; i < n; i++) {            unordered_map<int, int> dist;            for (int j = 0; j < n; j++) {                if (i == j)                    continue;                ++dist[distance(points[i], points[j])];            }            for (auto p : dist) {                if (p.second > 1)                    counts += p.second*(p.second - 1);            }        }        return counts;    }};