国产探花免费观看_亚洲丰满少妇自慰呻吟_97日韩有码在线_资源在线日韩欧美_一区二区精品毛片,辰东完美世界有声小说,欢乐颂第一季,yy玄幻小说排行榜完本

首頁 > 學院 > 開發設計 > 正文

LEETCODE--Number of Boomerangs

2019-11-11 05:44:21
字體:
來源:轉載
供稿:網友

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]]

熟悉使用map

class Solution {public: int numberOfBoomerangs(vector<pair<int, int>>& points) { int num = 0; for(int i = 0; i < points.size(); i++){ map<long,int> group; for(int j = 0; j < points.size(); j++){ if(i == j) continue; int distance_x = points[j].first - points[i].first; int distance_y = points[j].second - points[i].second; int key = distance_x * distance_x + distance_y * distance_y; ++group[key]; } for(auto& p : group){ int val = p.second; if(val > 1) num += val * (val - 1); } } return num; }};
發表評論 共有條評論
用戶名: 密碼:
驗證碼: 匿名發表
主站蜘蛛池模板: 金华市| 秭归县| 尉氏县| 甘南县| 罗江县| 赤峰市| 大渡口区| 浪卡子县| 雷波县| 班玛县| 景泰县| 贺州市| 息烽县| 南陵县| 高淳县| 郁南县| 东阿县| 通山县| 松江区| 根河市| 鲁山县| 喜德县| 静海县| 江津市| 谢通门县| 宝丰县| 佛山市| 余庆县| 麦盖提县| 连云港市| 乐陵市| 鄂州市| 北辰区| 留坝县| 永德县| 五大连池市| 泸西县| 常州市| 邯郸县| 平遥县| 平遥县|