leetcode刷题 447. Number of Boomerangs

最新推荐文章于 2021-09-15 00:03:55 发布

原创最新推荐文章于 2021-09-15 00:03:55 发布 · 370 阅读

0 ·

CC 4.0 BY-SA版权

leetcode刷题C++ 专栏收录该内容

43 篇文章

订阅专栏

本文介绍了一种计算平面内特定回旋镖数量的方法，使用C++实现。通过遍历每一点并统计与之距离相同的其他点来找出所有可能的回旋镖组合。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 andj 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]]

class Solution {
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points) {
        int len = points.size();
        int ret = 0;
        for(int i = 0; i < len; ++i)
        {
            map<int,int> res;
            for(int j = 0; j < len; ++j)
            {
                int tmp = dis(points[i],points[j]);
                res[tmp]++;
            }
            for(map<int,int>::iterator it = res.begin(); it != res.end(); ++it)
            {
                if(it -> second >= 2)
                    ret = ret + it -> second * (it-> second - 1);
            }
        }
        return ret;
    }
    int dis(pair<int,int> a,pair<int,int> b)
    {
        int x = pow(a.first - b.first,2);
        int y = pow(a.second - b.second,2);
        return x + y;
    }
};

核心思想：

1.用map记录下同距离的点数超过2的所有，然后对内部这些点进行排列比如a 跟b c d 的距离都是相等的

那么会有

a,b a.c

a,b a.d

a.c a.d

然后再调换顺序乘以2即可其实最终就是等差数列求和乘以2 就是(n+1)*n /2 * 2 也就是n*(n + 1)