有一个整数n,写一个函数f(n),返回0到n之间出现的"1"的个数

拓展延伸：

有一个整数n,写一个函数f(n，k),返回0到n之间出现的"k"的个数，0<=k<10

 

my codes as below:

 

    class Program

    {

        static void Main(string[] args)

        {

            //Test samples:                                              //Expected results:

            Console.WriteLine(TotalCountForNumbersSinceZero(-20,1));     //12

            Console.WriteLine(TotalCountForNumbersSinceZero(20, 1));     //12

            Console.WriteLine(TotalCountForNumbersSinceZero(0, 1));      //0

            Console.WriteLine(TotalCountForNumbersSinceZero(-20, 0));    //3

            Console.WriteLine(TotalCountForNumbersSinceZero(20, 0));     //3

            Console.WriteLine(TotalCountForNumbersSinceZero(0, 0));      //1

            Console.WriteLine(TotalCountForNumbersSinceZero(20, 11));    //0

 

            Console.ReadKey();

        }

        /// <summary>

        /// To count the appearance times of x in numbers from zero to lastNumber

        /// </summary>

        /// <param name="lastNumber"></param>

        /// <param name="x"></param>

        /// <returns></returns>

        static int TotalCountForNumbersSinceZero(int lastNumber, int x)

        {

            if (lastNumber < 0)

                lastNumber = -lastNumber;

            int totalCount = 0;

            for (int i = 0; i <= lastNumber; i++)

            {

                totalCount += CountForSingleNumber(i, x);

            }

            return totalCount;

        }

 

        /// <summary>

        /// To count the appearance times of x in number

        /// </summary>

        static int CountForSingleNumber(int number, int x)

        {

            if (x > 9 || x < 0)

            {

                return 0;

            }

            int count = 0;

            do

            {

                if (number % 10 == x)

                {

                    count++;

                }

                number = (number - number % 10) / 10;

            }

            while (number != 0);

            return count;

        }

    }