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).(在平面内给定n个点,”boomerang”是指由(i,j,k)组成的点元组,并且i和j之间的距离与i和k的距离相等)
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].
(找出”boomerang”的数量,假设n至多为500,并且点的坐标范围在[-10000,10000])
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]]
1.个人分析
题意是在一个含有n个二维点集合中找出符合指定条件的所有序列,该条件是一个点到其他两点的距离相等。
2.参考解法
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 dx = points[i].first - points[j].first;
int dy = points[i].second - points[j].second;
++m[dx*dx + dy*dy];
}
for(auto it = m.begin(); it != m.end(); ++it)
res += it->second*(it->second-1);
}
return res;
}
该解法的思路是,首先将所有点到其他所有点的距离的平方作为key保存到map容器中,key相同就累加value,然后计算每个元素的value*(value-1),并累加到返回结果中,最终就得到符合条件的点序列数。
3.总结
没有想出解法的原因是自己没有正确理解题意,可能即使想到与参考解法一样的思路,也很难想到将最终结果与value*(value-1)进行累加。