C#实现斐波那契数列的几种方法整理

什么是斐波那契数列？经典数学问题之一；斐波那契数列，又称黄金分割数列，指的是这样一个数列：1、1、2、3、5、8、13、21、……想必看到这个数列大家很容易的就推算出来后面好几项的值，那么到底有什么规律，简单说，就是前两项的和是第三项的值，用递归算法计第50位多少。

这个数列从第3项开始，每一项都等于前两项之和。

斐波那契数列：｛1,1,2,3,5,8,13,21...｝

递归算法，耗时最长的算法，效率很低。

public static long CalcA(int n)
{
  if (n <= 0) return 0;
  if (n <= 2) return 1;
  return checked(CalcA(n - 2) + CalcA(n - 1));
}

通过循环来实现

public static long CalcB(int n)
{
  if (n <= 0) return 0;
  var a = 1L;
  var b = 1L;
  var result = 1L;
  for (var i = 3; i <= n; i++)
  {
    result = checked(a + b);
    a = b;
    b = result;
  }
  return result;
}

通过循环的改进写法

public static long CalcC(int n)
{
  if (n <= 0) return 0;
  var a = 1L;
  var b = 1L;
  for (var i = 3; i <= n; i++)
  {
    b = checked(a + b);
    a = b - a;
  }
  return b;
}

通用公式法

/// <summary>
/// F(n)=(1/√5)*{[(1+√5)/2]^n - [(1-√5)/2]^n}
/// </summary>
/// <param name="n"></param>
/// <returns></returns>
public static long CalcD(int n)
{
  if (n <= 0) return 0;
  if (n <= 2) return 1; //加上，可减少运算。
  var a = 1 / Math.Sqrt(5);
  var b = Math.Pow((1 + Math.Sqrt(5)) / 2, n);
  var c = Math.Pow((1 - Math.Sqrt(5)) / 2, n);
  return checked((long)(a * (b - c)));
}

其他方法

using System;
using System.Diagnostics;


namespace Fibonacci
{
  class Program
  {
    static void Main(string[] args)
    {
      ulong result;

      int number = 10;
      Console.WriteLine("************* number={0} *************", number);

      Stopwatch watch1 = new Stopwatch();
      watch1.Start();
      result = F1(number);
      watch1.Stop();
      Console.WriteLine("F1({0})=" + result + " 耗时：" + watch1.Elapsed, number);

      Stopwatch watch2 = new Stopwatch();
      watch2.Start();
      result = F2(number);
      watch2.Stop();
      Console.WriteLine("F2({0})=" + result + " 耗时：" + watch2.Elapsed, number);

      Stopwatch watch3 = new Stopwatch();
      watch3.Start();
      result = F3(number);
      watch3.Stop();
      Console.WriteLine("F3({0})=" + result + " 耗时：" + watch3.Elapsed, number);

      Stopwatch watch4 = new Stopwatch();
      watch4.Start();
      double result4 = F4(number);
      watch4.Stop();
      Console.WriteLine("F4({0})=" + result4 + " 耗时：" + watch4.Elapsed, number);

      Console.WriteLine();

      Console.WriteLine("结束");
      Console.ReadKey();
    }

    /// <summary>
    /// 迭代法
    /// </summary>
    /// <param name="number"></param>
    /// <returns></returns>
    private static ulong F1(int number)
    {
      if (number == 1 || number == 2)
      {
        return 1;
      }
      else
      {
        return F1(number - 1) + F1(number - 2);
      }
      
    }

    /// <summary>
    /// 直接法
    /// </summary>
    /// <param name="number"></param>
    /// <returns></returns>
    private static ulong F2(int number)
    {
      ulong a = 1, b = 1;
      if (number == 1 || number == 2)
      {
        return 1;
      }
      else
      {
        for (int i = 3; i <= number; i++)
        {
          ulong c = a + b;
          b = a;
          a = c;
        }
        return a;
      }
    }

    /// <summary>
    /// 矩阵法
    /// </summary>
    /// <param name="n"></param>
    /// <returns></returns>
    static ulong F3(int n)
    {
      ulong[,] a = new ulong[2, 2] { { 1, 1 }, { 1, 0 } };
      ulong[,] b = MatirxPower(a, n);
      return b[1, 0];
    }

    #region F3
    static ulong[,] MatirxPower(ulong[,] a, int n)
    {
      if (n == 1) { return a; }
      else if (n == 2) { return MatirxMultiplication(a, a); }
      else if (n % 2 == 0)
      {
        ulong[,] temp = MatirxPower(a, n / 2);
        return MatirxMultiplication(temp, temp);
      }
      else
      {
        ulong[,] temp = MatirxPower(a, n / 2);
        return MatirxMultiplication(MatirxMultiplication(temp, temp), a);
      }
    }

    static ulong[,] MatirxMultiplication(ulong[,] a, ulong[,] b)
    {
      ulong[,] c = new ulong[2, 2];
      for (int i = 0; i < 2; i++)
      {
        for (int j = 0; j < 2; j++)
        {
          for (int k = 0; k < 2; k++)
          {
            c[i, j] += a[i, k] * b[k, j];
          }
        }
      }
      return c;
    }
    #endregion

    /// <summary>
    /// 通项公式法
    /// </summary>
    /// <param name="n"></param>
    /// <returns></returns>
    static double F4(int n)
    {
      double sqrt5 = Math.Sqrt(5);
      return (1/sqrt5*(Math.Pow((1+sqrt5)/2,n)-Math.Pow((1-sqrt5)/2,n)));
    }
  }
}

OK，就这些了。用的long类型来存储结果，当n>92时会内存溢出。

以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持易盾网络。