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 = ;
while (n != )
{
sum *= n;
n = n - ;
}
return sum;
} public int NumberOfBoomerangs(int[,] points)
{
var pointcount = points.GetLength();//点个数
var dim = points.GetLength();//每组个点的坐标维数
List<KeyValuePair<int, int>> list = new List<KeyValuePair<int, int>>(); for (int i = ; i < pointcount; i++)
{
var point = new KeyValuePair<int, int>(points[i, ], points[i, ]);
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 = ; 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, );
}
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, );
}
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, );
}
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, );
}
else
{
//dic[otherPoint][dis].Add(list[i]); dic[otherPoint][dis]++;
}
} #endregion 从点处理
}
}
} var sum = ; foreach (var d in dic)
{
foreach (var d2 in d.Value)
{
if (d2.Value > )
{
var ct = d2.Value;
//计算从Count中任取2个的排列数即A(count)(2)
sum += ct * (ct - );
}
}
} return sum;
}
}

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

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