小明数学考了优良中差？——二分法在区间统计的应用

前提：我的区间是用[,)计算的，集合已经升序排列好了。

        /// <summary>
        /// 
        /// </summary>
        /// <param name="data"></param>
        /// <param name="list"></param>
        /// <param name="low"></param>
        /// <param name="high"></param>
        /// <param name="left"></param>
        /// <param name="right"></param>
        /// <returns></returns>
        public static Tuple<int, int> BinarySearchRecursive(int data, IEnumerable<int> list, int low, int high, Tuple<int, int> left, Tuple<int, int> right)
        {

            int mid = (high + low) / 2;
            if (data >= left.Item2 && data <= right.Item2 && (right.Item1 - left.Item1 == 1))
            {
                return new Tuple<int, int>(left.Item2, right.Item2);
            }
            if (data >= list.ElementAt(list.Count() - 1))
            {
                //右边界
                return new Tuple<int, int>(list.ElementAt(list.Count() - 1), int.MaxValue);
            }
            if (data < list.ElementAt(0))
            {
                //左边界
                return new Tuple<int, int>(int.MinValue, list.ElementAt(0));
            }
            if (list.ElementAt(mid) > data)
            {
                return BinarySearchRecursive(data, list, low, mid - 1, left, new Tuple<int, int>(mid, list.ElementAt(mid)));
            }
            else if (list.ElementAt(mid) < data)
            {
                return BinarySearchRecursive(data, list, mid + 1, high, new Tuple<int, int>(mid, list.ElementAt(mid)), right);
            }
            //恰好和数组元素匹配的处理
            return new Tuple<int, int>(list.ElementAt(mid), list.ElementAt(mid + 1));
        }

有点恶心。不过测试通过就好啦。得到区间后其余就容易了，不过这个方法调用的时候有特别的技巧。那是作为递归结束的条件而特别那样的。

        static void Main(string[] args)
        {

            //var list = new List<int>() { 0, 50, 60, 80, 100 };
            var list = new List<int>() { 0, 50, 60, 800, 999 };
            Random r = new Random();

            int s = -50;
            var v = Class1.BinarySearchRecursive(s, list, 0, list.Count, new Tuple<int, int>(1, int.MaxValue), new Tuple<int, int>(0, int.MinValue));
            Console.WriteLine("随机数:{0}区间[{1},{2})", s, v.Item1, v.Item2);

            s = 0;
            v = Class1.BinarySearchRecursive(s, list, 0, list.Count, new Tuple<int, int>(1, int.MaxValue), new Tuple<int, int>(0, int.MinValue));
            Console.WriteLine("随机数:{0}区间[{1},{2})", s, v.Item1, v.Item2);

            s = 50;
            v = Class1.BinarySearchRecursive(s, list, 0, list.Count, new Tuple<int, int>(1, int.MaxValue), new Tuple<int, int>(0, int.MinValue));
            Console.WriteLine("随机数:{0}区间[{1},{2})", s, v.Item1, v.Item2);

            s = 88;
            v = Class1.BinarySearchRecursive(s, list, 0, list.Count, new Tuple<int, int>(1, int.MaxValue), new Tuple<int, int>(0, int.MinValue));
            Console.WriteLine("随机数:{0}区间[{1},{2})", s, v.Item1, v.Item2);

            s = 999;
            v = Class1.BinarySearchRecursive(s, list, 0, list.Count, new Tuple<int, int>(1, int.MaxValue), new Tuple<int, int>(0, int.MinValue));
            Console.WriteLine("随机数:{0}区间[{1},{2})", s, v.Item1, v.Item2);

            s = 1000;
            v = Class1.BinarySearchRecursive(s, list, 0, list.Count, new Tuple<int, int>(1, int.MaxValue), new Tuple<int, int>(0, int.MinValue));
            Console.WriteLine("随机数:{0}区间[{1},{2})", s, v.Item1, v.Item2);
            for (int i = 0; i < 50; i++)
            {
                var a = r.Next(-50, 1200);
                v = Class1.BinarySearchRecursive(a, list, 0, list.Count, new Tuple<int, int>(1, int.MaxValue), new Tuple<int, int>(0, int.MinValue));
                Console.WriteLine("随机数:{0}区间[{1},{2})", a, v.Item1, v.Item2);
            }
            Console.ReadKey();
        }

　　写的比较水，但是总结一下递归的要点应该就是把一些结果作为参数回传自身，收束的条件能够清晰的话，那问题就迎刃而解了。在这里我收束的条件是：这个数值>left并且<right并且left和right索引的差为1