LeetCode——Number of Boomerangs

LeetCode——Number of Boomerangs

Question

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]]

解题思路

注意这道题只需要求有多少对，没有具体要求是哪些，所以可以只记录点之间的距离。可以考虑用一个hash table记录每一个点到其他所有点的距离，如果相同距离数目大于等于2， 那么必有满足要求的组合。就是最后的时候注意如何求这样的组合。假如距离为6的点对是a， 那么这样的对数就是： (a * (a - 1) / 2) * 2.

具体实现

#include <iostream>
#include <vector>
#include <map>
using namespace std;
class Solution {
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points) {
        int res = 0;
        for (int i = 0; i < points.size(); i++) {
            map<int, int> dis_map;
            for (int j = 0; j < points.size(); j++) {
                if (i != j) {
                    int dis = distance(points[i], points[j]);
                    dis_map[dis]++;
                }
            }
            for (map<int, int>::iterator it = dis_map.begin(); it != dis_map.end(); it++) {
                res += (it->second * (it->second - 1) / 2) * 2;
            }
        }
        return res;
    }
    int distance(pair<int, int>& a, pair<int, int>& b) {
        int x = (a.first - b.first) * (a.first - b.first);
        int y = (a.second - b.second) * (a.second - b.second);
        return x + y;
    }
};