LeetCode 447 Number of Boomerangs
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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:2Explanation:The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]]题目链接
题意:
给定平面上所有两两不同的n个点,“boomerang”是一个点(i,j,k)的元组,使得i和j之间的距离等于i和k之间的距离(按照元组的顺序)。编写函数求boomerang的数量。
n至多为500,点的坐标都在10000, 10000(包括)范围内。
对于每一个boomerang,都存在一个中心点,即i点,枚举i点,记录i到其他个点的距离,看是否有相同的距离,假如有多个,则在其中选两个,构成组合数,2 * C(2, n),n为距离相等的数量,对于每一组,顺序都可以反转,所以需要乘两倍。
代码如下:
class Solution {public: int numberOfBoomerangs(vector<pair<int, int>>& points) { int ans = 0; for (int coor = 0; coor < points.size(); coor++) { map<long, int> dic; for (int i = 0; i < points.size(); i ++) { if (i == coor) continue; int dx = points[coor].first - points[i].first; int dy = points[coor].second - points[i].second; dic[dx*dx + dy*dy] ++; } for (auto &p : dic) { if (p.second > 1) { ans += p.second * (p.second - 1); } } } return ans; }};
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