边看边写(二)

    #region 用两个栈实现队列的功能    class CQueue<T> where T : IComparable<T>    {        private Stack<T> enterQ = new Stack<T>();        private Stack<T> outQ = new Stack<T>();        public void AppendTail(T  data)        {            enterQ.Push(data);        }        public T DeleteTail()        {            T outData;            if (outQ.Count()== 0)            {                while (enterQ.Count() > 0)                {                    outQ.Push(enterQ.Pop());                }            }            if (outQ.Count == 0)                outData = default(T);            else                outData = outQ.Pop();            return outData;        }    }    #endregion  #region 数组的整数次方        public bool DoubleEqual(double num1,double num2)        {            if ((num1 - num2) < 0.0000001 && (num1 - num2) > -0.0000001)            {                return true;            }            else                return false;        }        public double Power(double myBase, int exponent)        {             double result=1.0;             try             {                 if (DoubleEqual(myBase,0.0) == true)//注意1,double的数因为精度问题，不能直接比较                 {                     throw new Exception("基数不能为0！");                 }                 if (exponent < 0)//考虑exponent问负数的情况                 {                     myBase = 1 / myBase;                     exponent = -exponent;                 }                 result = PowerWithUnsignedExponent(myBase, (uint)exponent);                 Console.WriteLine(result);             }             catch (Exception ex)//异常处理             {                 Console.WriteLine(ex.Message);             }            return result;        }        //a^n=a^(n/2)*a^(n/2)(n为偶数)，a^n=a^(n/2-1)*a^(n/2-1)*a(n为奇数数)        public double PowerWithUnsignedExponent(double myBase, uint exponent)        {            if (exponent == 0)                return 1;            if (exponent == 1)                return myBase;            double result = PowerWithUnsignedExponent(myBase, exponent >> 1);//a^(n-1)=a^(n/2-1)*a^(n/2-1)            result *= result;            if ((exponent & 0x1) == 1)                result *= myBase;//a^n = a^(n-1)*a            return result;        }        #endregion