# 447.Number of Boomerangs

**447.Number of Boomerangs**

难度:Easy

> 给定平面上 n 对不同的点，“回旋镖” 是由点表示的元组 (i, j, k) ，其中 i 和 j 之间的距离和 i 和 k 之间的距离相等（需要考虑元组的顺序）。

找到所有回旋镖的数量。你可以假设 n 最大为 500，所有点的坐标在闭区间 \[-10000, 10000] 中。

```
示例:

输入:
[[0,0],[1,0],[2,0]]

输出:
2

解释:
两个回旋镖为 [[1,0],[0,0],[2,0]] 和 [[1,0],[2,0],[0,0]]
```

建立一个表示距离关系的矩阵，然后统计每行相同距离的点。

```
class Solution {
public:
    int numberOfBoomerangs(vector<vector<int>>& points) {
        int N=points.size();
        vector<vector<int>> distance(N, vector<int>(N,0));
        for(int i=0;i<N;i++)
            for(int j=i+1;j<N;j++)
            {
                distance[j][i]=distance[i][j]=pow((points[i][0]-points[j][0]),2)+pow((points[i][1]-points[j][1]),2);
                //cout<< distance[i][j]<<endl;
            }
        int sum=0;
        for(int i=0;i<N;i++)
        {
            unordered_map<int,int>tmp;
            for(int j=0;j<N;j++)
            {
                if(tmp.count(distance[i][j])) 
                   tmp[distance[i][j]]++ ;
                else 
                   tmp[distance[i][j]]=1;
            }
            for(auto iter=tmp.begin();iter!= tmp.end();iter++)
                   if(iter->second >1)
                   sum+=(iter->second)*(iter->second -1);
        }
        return sum;
    }
};
```


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