447. Number of Boomerangs

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 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:2Explanation:The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]]

思路：

本题思路很简单，实现两个函数，一个是求距离的函数

sumq(pair<int,int> p1,pair<int,int> p2)

一个是求个数的函数

sumn(map<long,int> res)

通过两个嵌套for循环来实现遍历

代码：

class Solution {public:    int numberOfBoomerangs(vector<pair<int, int>>& points) {        map<long,int> res;        int sum = 0;        for(int i =0;i<=points.size()-1;i++)        {            for(int j =0;j<=points.size()-1;j++)            {                if(j!=i)                    res[(sumq(points[i],points[j]))]++;            }            sum+=sumn(res);            res.clear();                }    return sum;            }private:    long sumq(pair<int,int> p1,pair<int,int> p2){        long distance = (p1.first-p2.first)*(p1.first-p2.first)+(p1.second-p2.second)*(p1.second-p2.second);        return distance;    }    int sumn(map<long,int> res){        map<long,int>::iterator itr = res.begin();        int count =0;        while(itr!=res.end())        {            if(itr->second>1)                count+=itr->second*(itr->second-1);            itr++;        }        return count;            }    };