Number of Boomerangs:
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]]
Sulotion:
#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;
// 时间复杂度:o(n^2)
// 空间复杂度:o(n)
class Solution {
public:
int numberOfBoomerangs(vector<pair<int, int>>& points) {
int res = 0;
for( int i = 0; i < points.size(); i++ ){
unordered_map<int,int> record;
for( int j = 0; j < points.size(); j++ ){
if( j != i ){
record[ dis(points[i], points[j]) ] ++;
}
}
for( unordered_map<int,int>::iterator iter = record.begin();
iter != record.end(); iter++)
res += (iter->second) * (iter->second - 1);
}
return res;
}
private:
int dis( const pair<int,int> &pa, const pair<int,int> &pb ){
return (pa.first - pb.first) * (pa.first - pb.first) +
(pa.second - pb.second) * (pa.second - pb.second);
}
};
总结: 这也是一个灵活应用键值对的例子,这里的map记录的是距离和这个距离的频次。小技巧是把i作为“枢纽”。