LeetCode 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 iand 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]]
给定一堆点集，问回飞镖的个数。一个回飞镖是一个三点集(i,j,k)，且满足i到j和k的距离相等。
如果和i距离相等的点有n个，则这样的三点集有，也就是从n个点中拿两个点做排列。所以问题就转换为对每个点，都求其他所有点和该点的距离，求到距离相等的点的个数n，然后代入公式计算。
代码如下：
[cpp]
class Solution {
public:
int numberOfBoomerangs(vector<pair<int, int>>& points) {
int ans = 0;
for (int i = 0; i < points.size(); ++i) {
unordered_map<int, int> dist_cnt;
for (int j = 0; j < points.size(); ++j) {
int xdiff = points[i].first – points[j].first;
int ydiff = points[i].second – points[j].second;
++dist_cnt[xdiff*xdiff + ydiff*ydiff];
}
for (auto it = dist_cnt.begin(); it != dist_cnt.end(); ++it)ans += it->second*(it->second – 1);
}
return ans;
}
};
[/cpp]
本代码提交AC，用时369MS。
因为题中说到所有点都是不相同的，所以第7行没必要限制j!=i，即使j==i，算出来的距离是0，dist_cnt[0]=1，也就是只有点x一个点和自己的距离是0。到第12行计算排列数时，n(n-1)=0，所以没关系。]]>