[Leetcode] 447. Number of Boomerangs 解题报告

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).

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]]

思路：

对于每个点，计算出来它和其它所有点的距离。对于每个距离，假设有m个点，则可以构成m * (m - 1)个Boomerangs，返回所有结果之和即可。思路挺简单。

代码：

class Solution {
public:
int numberOfBoomerangs(vector<pair<int, int>>& points) {
int ret = 0;
for(int i = 0; i < points.size(); ++i) {
unordered_map<long, int> hash;
for(int j = 0; j < points.size(); ++j) {
if(j == i) {
continue;
}
long dist = (points[i].first - points[j].first) * (points[i].first - points[j].first);
dist += (points[i].second - points[j].second) * (points[i].second - points[j].second);
++hash[dist];
}
for(auto val : hash) {
ret += val.second * (val.second - 1);
}
}
return ret;
}
};