LeetCode-Easy部分标签为HashTable 447. Number of Boomerangs

原题

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]]

题目分析

给出n组点，找出所有boomerang 的个数，boomerang 定义： d(Pi,Pj)=d(Pi,Pk)。题目涉及到三个点，先保持一个轴点rot，然后找与其他点的距离，若个数为m，m满足m>1，则找到轴点rot 的Boomerangs 个数为A(m,2) ，继续遍历其他轴点，返回累加和。

代码分析

        public static int NumberOfBoomerangs(int[,] points)        {            Dictionary<double, int> dict = new Dictionary<double, int>();            int len = points.GetUpperBound(0);            int rtnCnt=0;            for (int i = 0; i <= len; i++)            {                //3点变2点                for (int j = 0; j <= len; j++)                {                    if (i == j) continue;                    double d = distance(points[i, 0], points[j, 0], points[i, 1], points[j, 1]);                    if (dict.ContainsKey(d))                        dict[d]++;                    else dict.Add(d, 1);                }                foreach(var item in dict)                {                    if (item.Value > 1)                    {                        //如果找到了value个，因为有顺序，所以排序                        rtnCnt += item.Value*(item.Value-1);                     }                }                dict.Clear();            }            return rtnCnt;        }        private static double distance(int x1, int x2, int y1, int y2)        {            int x = x1 - x2;            int y = y1 - y2;            return Math.Sqrt(x * x + y * y);        }

结果分析

算法运行时间为 Runtime: 372 ms，排名85% 。
时间复杂度O(n^2)，控件复杂度O(n)。

更多参考

LeetCode-Easy部分中标签为HashTable的所有题目