leetcode447

public class Solution
    {
        /// <summary>
        /// 计算两个点的距离
        /// </summary>
        /// <param name="x1"></param>
        /// <param name="y1"></param>
        /// <param name="x2"></param>
        /// <param name="y2"></param>
        /// <returns></returns>
        private double getDistance(int x1, int y1, int x2, int y2)
        {
            var dis = (y2 - y1) * (y2 - y1) + (x2 - x1) * (x2 - x1);
            return dis;
        }

        /// <summary>
        /// 计算n的阶乘
        /// </summary>
        /// <param name="n"></param>
        /// <returns></returns>
        private int SetpNum(int n)
        {
            int sum = 1;
            while (n != 1)
            {
                sum *= n;
                n = n - 1;
            }
            return sum;
        }

        public int NumberOfBoomerangs(int[,] points)
        {
            var pointcount = points.GetLength(0);//点个数
            var dim = points.GetLength(1);//每组个点的坐标维数
            List<KeyValuePair<int, int>> list = new List<KeyValuePair<int, int>>();

            for (int i = 0; i < pointcount; i++)
            {
                var point = new KeyValuePair<int, int>(points[i, 0], points[i, 1]);
                list.Add(point);
            }

            //var dic = new Dictionary<KeyValuePair<int, int>, Dictionary<double, List<KeyValuePair<int, int>>>>();
            var dic = new Dictionary<KeyValuePair<int, int>, Dictionary<double, int>>();

            for (int i = 0; i < list.Count; i++)
            {
                for (int j = i; j < list.Count; j++)
                {
                    if (i != j)
                    {
                        var curentPoint = list[i];//主点
                        var otherPoint = list[j];//从点

                        var dis = getDistance(list[i].Key, list[i].Value, list[j].Key, list[j].Value);//dis值是同一个

                        #region 主点处理
                        if (!dic.ContainsKey(curentPoint))
                        {
                            //dic.Add(curentPoint, new Dictionary<double, List<KeyValuePair<int, int>>>());
                            //dic[curentPoint].Add(dis, new List<KeyValuePair<int, int>>());
                            //dic[curentPoint][dis].Add(list[j]);

                            dic.Add(curentPoint, new Dictionary<double, int>());
                            dic[curentPoint].Add(dis, 1);
                        }
                        else
                        {
                            if (!dic[curentPoint].ContainsKey(dis))
                            {
                                //dic[curentPoint].Add(dis, new List<KeyValuePair<int, int>>());
                                //dic[curentPoint][dis].Add(list[j]);

                                dic[curentPoint].Add(dis, 1);
                            }
                            else
                            {
                                //dic[curentPoint][dis].Add(list[j]);

                                dic[curentPoint][dis]++;
                            }
                        }
                        #endregion 主点处理

                        #region 从点处理
                        if (!dic.ContainsKey(otherPoint))
                        {
                            //dic.Add(otherPoint, new Dictionary<double, List<KeyValuePair<int, int>>>());
                            //dic[otherPoint].Add(dis, new List<KeyValuePair<int, int>>());
                            //dic[otherPoint][dis].Add(list[i]);

                            dic.Add(otherPoint, new Dictionary<double, int>());
                            dic[otherPoint].Add(dis, 1);
                        }
                        else
                        {
                            if (!dic[otherPoint].ContainsKey(dis))
                            {
                                //dic[otherPoint].Add(dis, new List<KeyValuePair<int, int>>());
                                //dic[otherPoint][dis].Add(list[i]);

                                dic[otherPoint].Add(dis, 1);
                            }
                            else
                            {
                                //dic[otherPoint][dis].Add(list[i]);

                                dic[otherPoint][dis]++;
                            }
                        }

                        #endregion 从点处理
                    }
                }
            }

            var sum = 0;

            foreach (var d in dic)
            {
                foreach (var d2 in d.Value)
                {
                    if (d2.Value > 1)
                    {
                        var ct = d2.Value;
                        //计算从Count中任取2个的排列数即A(count)(2)
                        sum += ct * (ct - 1);
                    }
                }
            }

            return sum;
        }
    }

https://leetcode.com/problems/number-of-boomerangs/#/description