斐波那契数列程序

斐波那契数是以下整数序列中的数字。
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …..

在数学术语中，斐波那契数列 Fn 由递推关系定义

Fn = Fn-1 + Fn-2

带有种子值

F0 = 0 and F1 = 1.

给定一个数 n，打印第 n 个斐波那契数。

例子：

Input  : n = 2
Output : 1

Input  : n = 9
Output : 34

编写一个返回 F _n的函数int fib(int n) 。例如，如果n = 0，那么fib()应该返回 0。如果 n = 1，那么它应该返回 1。对于 n > 1，它应该返回 F _n-1 + F _n-2

For n = 9
Output:34

以下是获取第 n 个斐波那契数的不同方法。

方法一(使用递归)
一种简单的方法，即上面给出的直接递归实现数学递推关系。

C++

//Fibonacci Series using Recursion
#include
using namespace std;
 
int fib(int n)
{
    if (n <= 1)
        return n;
    return fib(n-1) + fib(n-2);
}
 
int main ()
{
    int n = 9;
    cout << fib(n);
    getchar();
    return 0;
}
 
// This code is contributed
// by Akanksha Rai

C

//Fibonacci Series using Recursion
#include
int fib(int n)
{
   if (n <= 1)
      return n;
   return fib(n-1) + fib(n-2);
}
 
int main ()
{
  int n = 9;
  printf("%d", fib(n));
  getchar();
  return 0;
}

Java

//Fibonacci Series using Recursion
class fibonacci
{
    static int fib(int n)
    {
    if (n <= 1)
       return n;
    return fib(n-1) + fib(n-2);
    }
      
    public static void main (String args[])
    {
    int n = 9;
    System.out.println(fib(n));
    }
}
/* This code is contributed by Rajat Mishra */

Python

# Function for nth Fibonacci number
 
def Fibonacci(n):
    if n<0:
        print("Incorrect input")
    # First Fibonacci number is 0
    elif n==0:
        return 0
    # Second Fibonacci number is 1
    elif n==1:
        return 1
    else:
        return Fibonacci(n-1)+Fibonacci(n-2)
 
# Driver Program
 
print(Fibonacci(9))
 
#This code is contributed by Saket Modi

C#

// C# program for Fibonacci Series
// using Recursion
using System;
 
public class GFG
{
    public static int Fib(int n)
    {
        if (n <= 1)
        {
            return n;
        }
        else
        {
            return Fib(n - 1) + Fib(n - 2);
        }
    }
         
    // driver code
    public static void Main(string[] args)
    {
        int n = 9;
        Console.Write(Fib(n));
    }
}
 
// This code is contributed by Sam007

Javascript

//Fibonacci Series using Recursion
    let n = 9;
     
    // function returns the Fibonacci number
    function fib(n) {
    if (n <= 1)
        return n;
    return fib(n-1) + fib(n-2);
}
  
    //function call
    document.write(fib(n));
//This code is contributed by Surbhi Tyagi

输出

34

时间复杂度： T(n) = T(n-1) + T(n-2) 这是指数。

我们可以观察到这个实现做了很多重复的工作(见下面的递归树)。所以这是第 n 个斐波那契数的糟糕实现。

fib(5)   
                     /                \
               fib(4)                fib(3)   
             /        \              /       \ 
         fib(3)      fib(2)         fib(2)   fib(1)
        /    \       /    \        /      \
  fib(2)   fib(1)  fib(1) fib(0) fib(1) fib(0)
  /     \
fib(1) fib(0)

额外空间：如果我们考虑函数调用堆栈大小，则为 O(n)，否则为 O(1)。

方法二(使用动态规划)
我们可以通过存储到目前为止计算的斐波那契数来避免方法 1 中所做的重复工作。

C++

// C++ program for Fibonacci Series 
// using Dynamic Programming
#include
using namespace std;
 
class GFG{
     
public:
int fib(int n)
{
     
    // Declare an array to store
    // Fibonacci numbers.
    // 1 extra to handle
    // case, n = 0
    int f[n + 2];
    int i;
 
    // 0th and 1st number of the
    // series are 0 and 1
    f[0] = 0;
    f[1] = 1;
 
    for(i = 2; i <= n; i++)
    {
         
       //Add the previous 2 numbers
       // in the series and store it
       f[i] = f[i - 1] + f[i - 2];
    }
    return f[n];
    }
};
 
// Driver code
int main ()
{
    GFG g;
    int n = 9;
     
    cout << g.fib(n);
    return 0;
}
 
// This code is contributed by SoumikMondal

C

//Fibonacci Series using Dynamic Programming
#include
 
int fib(int n)
{
  /* Declare an array to store Fibonacci numbers. */
  int f[n+2];   // 1 extra to handle case, n = 0
  int i;
 
  /* 0th and 1st number of the series are 0 and 1*/
  f[0] = 0;
  f[1] = 1;
 
  for (i = 2; i <= n; i++)
  {
      /* Add the previous 2 numbers in the series
         and store it */
      f[i] = f[i-1] + f[i-2];
  }
 
  return f[n];
}
 
int main ()
{
  int n = 9;
  printf("%d", fib(n));
  getchar();
  return 0;
}

Java

// Fibonacci Series using Dynamic Programming
class fibonacci
{
   static int fib(int n)
    {
    /* Declare an array to store Fibonacci numbers. */
    int f[] = new int[n+2]; // 1 extra to handle case, n = 0
    int i;
      
    /* 0th and 1st number of the series are 0 and 1*/
    f[0] = 0;
    f[1] = 1;
     
    for (i = 2; i <= n; i++)
    {
       /* Add the previous 2 numbers in the series
         and store it */
        f[i] = f[i-1] + f[i-2];
    }
      
    return f[n];
    }
      
    public static void main (String args[])
    {
        int n = 9;
        System.out.println(fib(n));
    }
}
/* This code is contributed by Rajat Mishra */

Python

# Fibonacci Series using Dynamic Programming
def fibonacci(n):
     
    # Taking 1st two fibonacci numbers as 0 and 1
    f = [0, 1]
     
     
    for i in range(2, n+1):
        f.append(f[i-1] + f[i-2])
    return f[n]
     
print(fibonacci(9))

C＃

// C# program for Fibonacci Series
// using Dynamic Programming
using System;
class fibonacci {
     
static int fib(int n)
    {
         
        // Declare an array to
        // store Fibonacci numbers.
        // 1 extra to handle
        // case, n = 0
        int []f = new int[n + 2];
        int i;
         
        /* 0th and 1st number of the
           series are 0 and 1 */
        f[0] = 0;
        f[1] = 1;
         
        for (i = 2; i <= n; i++)
        {
            /* Add the previous 2 numbers
               in the series and store it */
            f[i] = f[i - 1] + f[i - 2];
        }
         
        return f[n];
    }
     
    // Driver Code
    public static void Main ()
    {
        int n = 9;
        Console.WriteLine(fib(n));
    }
}
 
// This code is contributed by anuj_67.

Javascript

 
// Fibonacci Series using Dynamic Programming
 
    function  fib(n)
    {
        /* Declare an array to store Fibonacci numbers. */
        let f = new Array(n+2); // 1 extra to handle case, n = 0
        let i;
        /* 0th and 1st number of the series are 0 and 1*/
        f[0] = 0;
        f[1] = 1;
        for (i = 2; i <= n; i++)
        {
            /* Add the previous 2 numbers in the series
            and store it */
            f[i] = f[i-1] + f[i-2];
        }
        return f[n];
    }
    let n=9;
    document.write(fib(n));
     
    // This code is contributed by avanitrachhadiya2155
     

输出

34

方法三(空间优化方法二)
我们可以通过存储前两个数字来优化方法 2 中使用的空间，因为这就是我们获得下一个斐波那契数列所需的全部内容。

C++

// Fibonacci Series using Space Optimized Method
#include
using namespace std;
 
int fib(int n)
{
    int a = 0, b = 1, c, i;
    if( n == 0)
        return a;
    for(i = 2; i <= n; i++)
    {
       c = a + b;
       a = b;
       b = c;
    }
    return b;
}
 
// Driver code
int main()
{
    int n = 9;
     
    cout << fib(n);
    return 0;
}
 
// This code is contributed by Code_Mech

C

// Fibonacci Series using Space Optimized Method
#include
int fib(int n)
{
  int a = 0, b = 1, c, i;
  if( n == 0)
    return a;
  for (i = 2; i <= n; i++)
  {
     c = a + b;
     a = b;
     b = c;
  }
  return b;
}
 
int main ()
{
  int n = 9;
  printf("%d", fib(n));
  getchar();
  return 0;
}

Java

// Java program for Fibonacci Series using Space
// Optimized Method
class fibonacci
{
    static int fib(int n)
    {
        int a = 0, b = 1, c;
        if (n == 0)
            return a;
        for (int i = 2; i <= n; i++)
        {
            c = a + b;
            a = b;
            b = c;
        }
        return b;
    }
 
    public static void main (String args[])
    {
        int n = 9;
        System.out.println(fib(n));
    }
}
 
// This code is contributed by Mihir Joshi

Python

# Function for nth fibonacci number - Space Optimisation
# Taking 1st two fibonacci numbers as 0 and 1
 
def fibonacci(n):
    a = 0
    b = 1
    if n < 0:
        print("Incorrect input")
    elif n == 0:
        return a
    elif n == 1:
        return b
    else:
        for i in range(2,n+1):
            c = a + b
            a = b
            b = c
        return b
 
# Driver Program
 
print(fibonacci(9))
 
#This code is contributed by Saket Modi

C＃

// C# program for Fibonacci Series
// using Space Optimized Method
using System;
 
namespace Fib
{
    public class GFG
    {
        static int Fib(int n)
        {
            int a = 0, b = 1, c = 0;
             
            // To return the first Fibonacci number
            if (n == 0) return a;
     
            for (int i = 2; i <= n; i++)
            {
                c = a + b;
                a = b;
                b = c;
            }
     
            return b;
        }
         
    // Driver function
    public static void Main(string[] args)
        {
             
            int n = 9;
            Console.Write("{0} ", Fib(n));
        }
    }
}
 
// This code is contributed by Sam007.

Javascript

 
// Javascript program for Fibonacci Series using Space Optimized Method
 
function fib(n)
{
    let a = 0, b = 1, c, i;
    if( n == 0)
        return a;
    for(i = 2; i <= n; i++)
    {
    c = a + b;
    a = b;
    b = c;
    }
    return b;
}
 
// Driver code
 
    let n = 9;
     
    document.write(fib(n));
 
// This code is contributed by Mayank Tyagi
 

输出

34

时间复杂度： O(n)
额外空间： O(1)

方法 4(使用矩阵 {{1, 1}, {1, 0}} 的幂)
这是另一个 O(n)，它依赖于这样一个事实：如果我们将矩阵 M = {{1,1},{1,0}} 乘以 n 次(换句话说，计算 power(M, n))，那么我们得到第 (n+1) 个斐波那契数作为结果矩阵中第 (0, 0) 行和第 (0, 0) 列的元素。
矩阵表示给出了以下斐波那契数列的封闭表达式：

C++

#include
using namespace std;
 
// Helper function that multiplies 2
// matrices F and M of size 2*2, and
// puts the multiplication result
// back to F[][]
void multiply(int F[2][2], int M[2][2]);
 
// Helper function that calculates F[][]
// raise to the power n and puts the
// result in F[][]
// Note that this function is designed
// only for fib() and won't work as
// general power function
void power(int F[2][2], int n);
 
int fib(int n)
{
    int F[2][2] = { { 1, 1 }, { 1, 0 } };
     
    if (n == 0)
        return 0;
         
    power(F, n - 1);
     
    return F[0][0];
}
 
void multiply(int F[2][2], int M[2][2])
{
    int x = F[0][0] * M[0][0] +
            F[0][1] * M[1][0];
    int y = F[0][0] * M[0][1] +
            F[0][1] * M[1][1];
    int z = F[1][0] * M[0][0] +
            F[1][1] * M[1][0];
    int w = F[1][0] * M[0][1] +
            F[1][1] * M[1][1];
     
    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;
}
 
void power(int F[2][2], int n)
{
    int i;
    int M[2][2] = { { 1, 1 }, { 1, 0 } };
     
    // n - 1 times multiply the
    // matrix to {{1,0},{0,1}}
    for(i = 2; i <= n; i++)
        multiply(F, M);
}
 
// Driver code
int main()
{
    int n = 9;
     
    cout << " " <<  fib(n);
     
    return 0;
}
 
// This code is contributed by shivanisinghss2110

C

#include 
 
/* Helper function that multiplies 2 matrices F and M of size 2*2, and
  puts the multiplication result back to F[][] */
void multiply(int F[2][2], int M[2][2]);
 
/* Helper function that calculates F[][] raise to the power n and puts the
  result in F[][]
  Note that this function is designed only for fib() and won't work as general
  power function */
void power(int F[2][2], int n);
 
int fib(int n)
{
  int F[2][2] = {{1,1},{1,0}};
  if (n == 0)
      return 0;
  power(F, n-1);
 
  return F[0][0];
}
 
void multiply(int F[2][2], int M[2][2])
{
  int x =  F[0][0]*M[0][0] + F[0][1]*M[1][0];
  int y =  F[0][0]*M[0][1] + F[0][1]*M[1][1];
  int z =  F[1][0]*M[0][0] + F[1][1]*M[1][0];
  int w =  F[1][0]*M[0][1] + F[1][1]*M[1][1];
 
  F[0][0] = x;
  F[0][1] = y;
  F[1][0] = z;
  F[1][1] = w;
}
 
void power(int F[2][2], int n)
{
  int i;
  int M[2][2] = {{1,1},{1,0}};
 
  // n - 1 times multiply the matrix to {{1,0},{0,1}}
  for (i = 2; i <= n; i++)
      multiply(F, M);
}
 
/* Driver program to test above function */
int main()
{
  int n = 9;
  printf("%d", fib(n));
  getchar();
  return 0;
}

Java

class fibonacci
{
     
    static int fib(int n)
    {
    int F[][] = new int[][]{{1,1},{1,0}};
    if (n == 0)
        return 0;
    power(F, n-1);
     
       return F[0][0];
    }
      
     /* Helper function that multiplies 2 matrices F and M of size 2*2, and
     puts the multiplication result back to F[][] */
    static void multiply(int F[][], int M[][])
    {
    int x =  F[0][0]*M[0][0] + F[0][1]*M[1][0];
    int y =  F[0][0]*M[0][1] + F[0][1]*M[1][1];
    int z =  F[1][0]*M[0][0] + F[1][1]*M[1][0];
    int w =  F[1][0]*M[0][1] + F[1][1]*M[1][1];
      
    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;
    }
 
    /* Helper function that calculates F[][] raise to the power n and puts the
    result in F[][]
    Note that this function is designed only for fib() and won't work as general
    power function */
    static void power(int F[][], int n)
    {
    int i;
    int M[][] = new int[][]{{1,1},{1,0}};
     
    // n - 1 times multiply the matrix to {{1,0},{0,1}}
    for (i = 2; i <= n; i++)
        multiply(F, M);
    }
      
    /* Driver program to test above function */
    public static void main (String args[])
    {
    int n = 9;
    System.out.println(fib(n));
    }
}
/* This code is contributed by Rajat Mishra */

蟒蛇3

# Helper function that multiplies
# 2 matrices F and M of size 2*2,
# and puts the multiplication
# result back to F[][]
 
# Helper function that calculates
# F[][] raise to the power n and
# puts the result in F[][]
# Note that this function is
# designed only for fib() and
# won't work as general
# power function
def fib(n):
    F = [[1, 1],
         [1, 0]]
    if (n == 0):
        return 0
    power(F, n - 1)
     
    return F[0][0]
 
def multiply(F, M):
 
    x = (F[0][0] * M[0][0] +
         F[0][1] * M[1][0])
    y = (F[0][0] * M[0][1] +
         F[0][1] * M[1][1])
    z = (F[1][0] * M[0][0] +
         F[1][1] * M[1][0])
    w = (F[1][0] * M[0][1] +
         F[1][1] * M[1][1])
     
    F[0][0] = x
    F[0][1] = y
    F[1][0] = z
    F[1][1] = w
 
def power(F, n):
 
    M = [[1, 1],
         [1, 0]]
 
    # n - 1 times multiply the
    # matrix to {{1,0},{0,1}}
    for i in range(2, n + 1):
        multiply(F, M)
 
# Driver Code
if __name__ == "__main__":
    n = 9
    print(fib(n))
 
# This code is contributed
# by ChitraNayal

C＃

using System;
 
class GFG {
     
    static int fib(int n)
    {
        int [,]F = new int[,] {{1, 1},
                               {1, 0} };
        if (n == 0)
            return 0;
        power(F, n-1);
         
        return F[0,0];
    }
     
    /* Helper function that multiplies 2
    matrices F and M of size 2*2, and puts
    the multiplication result back to F[][] */
    static void multiply(int [,]F, int [,]M)
    {
        int x = F[0,0]*M[0,0] + F[0,1]*M[1,0];
        int y = F[0,0]*M[0,1] + F[0,1]*M[1,1];
        int z = F[1,0]*M[0,0] + F[1,1]*M[1,0];
        int w = F[1,0]*M[0,1] + F[1,1]*M[1,1];
         
        F[0,0] = x;
        F[0,1] = y;
        F[1,0] = z;
        F[1,1] = w;
    }
 
    /* Helper function that calculates F[][]
    raise to the power n and puts the result
    in F[][] Note that this function is designed
    only for fib() and won't work as general
    power function */
    static void power(int [,]F, int n)
    {
        int i;
        int [,]M = new int[,]{{1, 1},
                              {1, 0} };
         
        // n - 1 times multiply the matrix to
        // {{1,0},{0,1}}
        for (i = 2; i <= n; i++)
            multiply(F, M);
    }
     
    /* Driver program to test above function */
    public static void Main ()
    {
        int n = 9;
        Console.WriteLine(fib(n));
    }
}
 
// This code is contributed by anuj_67.

Javascript

 
// Note that this function is designed
// only for fib() and won't work as
// general power function
 
  
function fib( n)
{
     var F = [ [ 1, 1 ], [ 1, 0 ] ];
      
    if (n == 0)
        return 0;
          
    power(F, n - 1);
      
    return F[0][0];
}
// Helper function that multiplies 2
// matrices F and M of size 2*2, and
// puts the multiplication result
// back to F[][]
  function multiply( F, M )
{
    x = F[0][0] * M[0][0] +
            F[0][1] * M[1][0];
    y = F[0][0] * M[0][1] +
            F[0][1] * M[1][1];
     z = F[1][0] * M[0][0] +
            F[1][1] * M[1][0];
    w = F[1][0] * M[0][1] +
            F[1][1] * M[1][1];
      
    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;
}
// Helper function that calculates F[][]
// raise to the power n and puts the
// result in F[][]
  
function power( F, n)
{
    var i;
     var M = [[ 1, 1 ], [ 1, 0 ]];
      
    // n - 1 times multiply the
    // matrix to {{1,0},{0,1}}
    for(i = 2; i <= n; i++)
        multiply(F, M);
}
  
// Driver code
 
    var   n = 9;
      
     document.write (" " +  fib(n));
 //This code is contributed by sweetyty
 

输出

34

时间复杂度： O(n)
额外空间： O(1)

方法五(优化方法四)
方法 4 可以优化为 O(Logn) 时间复杂度。我们可以在之前的方法中进行递归乘法来获得 power(M, n) (类似于本文中所做的优化)

C++

// Fibonacci Series using Optimized Method
#include 
using namespace std;
 
void multiply(int F[2][2], int M[2][2]);
void power(int F[2][2], int n);
 
// Function that returns nth Fibonacci number
int fib(int n)
{
    int F[2][2] = {{1, 1}, {1, 0}};
    if (n == 0)
        return 0;
    power(F, n - 1);
 
    return F[0][0];
}
 
// Optimized version of power() in method 4
void power(int F[2][2], int n)
{
    if(n == 0 || n == 1)
       return;
    int M[2][2] = {{1, 1}, {1, 0}};
     
    power(F, n / 2);
    multiply(F, F);
     
    if (n % 2 != 0)
        multiply(F, M);
}
 
void multiply(int F[2][2], int M[2][2])
{
    int x = F[0][0] * M[0][0] + F[0][1] * M[1][0];
    int y = F[0][0] * M[0][1] + F[0][1] * M[1][1];
    int z = F[1][0] * M[0][0] + F[1][1] * M[1][0];
    int w = F[1][0] * M[0][1] + F[1][1] * M[1][1];
     
    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;
}
 
// Driver code
int main()
{
    int n = 9;
     
    cout << fib(9);
    getchar();
     
    return 0;
}
 
// This code is contributed by Nidhi_biet

C

#include 
 
void multiply(int F[2][2], int M[2][2]);
 
void power(int F[2][2], int n);
 
/* function that returns nth Fibonacci number */
int fib(int n)
{
  int F[2][2] = {{1,1},{1,0}};
  if (n == 0)
    return 0;
  power(F, n-1);
  return F[0][0];
}
 
/* Optimized version of power() in method 4 */
void power(int F[2][2], int n)
{
  if( n == 0 || n == 1)
      return;
  int M[2][2] = {{1,1},{1,0}};
 
  power(F, n/2);
  multiply(F, F);
 
  if (n%2 != 0)
     multiply(F, M);
}
 
void multiply(int F[2][2], int M[2][2])
{
  int x =  F[0][0]*M[0][0] + F[0][1]*M[1][0];
  int y =  F[0][0]*M[0][1] + F[0][1]*M[1][1];
  int z =  F[1][0]*M[0][0] + F[1][1]*M[1][0];
  int w =  F[1][0]*M[0][1] + F[1][1]*M[1][1];
 
  F[0][0] = x;
  F[0][1] = y;
  F[1][0] = z;
  F[1][1] = w;
}
 
/* Driver program to test above function */
int main()
{
  int n = 9;
  printf("%d", fib(9));
  getchar();
  return 0;
}

Java

//Fibonacci Series using Optimized Method
class fibonacci
{
    /* function that returns nth Fibonacci number */
    static int fib(int n)
    {
    int F[][] = new int[][]{{1,1},{1,0}};
    if (n == 0)
        return 0;
    power(F, n-1);
      
    return F[0][0];
    }
      
    static void multiply(int F[][], int M[][])
    {
    int x =  F[0][0]*M[0][0] + F[0][1]*M[1][0];
    int y =  F[0][0]*M[0][1] + F[0][1]*M[1][1];
    int z =  F[1][0]*M[0][0] + F[1][1]*M[1][0];
    int w =  F[1][0]*M[0][1] + F[1][1]*M[1][1];
     
    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;
    }
      
    /* Optimized version of power() in method 4 */
    static void power(int F[][], int n)
    {
    if( n == 0 || n == 1)
      return;
    int M[][] = new int[][]{{1,1},{1,0}};
      
    power(F, n/2);
    multiply(F, F);
      
    if (n%2 != 0)
       multiply(F, M);
    }
     
    /* Driver program to test above function */
    public static void main (String args[])
    {
         int n = 9;
     System.out.println(fib(n));
    }
}
/* This code is contributed by Rajat Mishra */

蟒蛇3

# Fibonacci Series using
# Optimized Method
 
# function that returns nth
# Fibonacci number
def fib(n):
     
    F = [[1, 1],
         [1, 0]]
    if (n == 0):
        return 0
    power(F, n - 1)
         
    return F[0][0]
     
def multiply(F, M):
     
    x = (F[0][0] * M[0][0] +
         F[0][1] * M[1][0])
    y = (F[0][0] * M[0][1] +
         F[0][1] * M[1][1])
    z = (F[1][0] * M[0][0] +
         F[1][1] * M[1][0])
    w = (F[1][0] * M[0][1] +
         F[1][1] * M[1][1])
     
    F[0][0] = x
    F[0][1] = y
    F[1][0] = z
    F[1][1] = w
         
# Optimized version of
# power() in method 4
def power(F, n):
 
    if( n == 0 or n == 1):
        return;
    M = [[1, 1],
         [1, 0]];
         
    power(F, n // 2)
    multiply(F, F)
         
    if (n % 2 != 0):
        multiply(F, M)
     
# Driver Code
if __name__ == "__main__":
    n = 9
    print(fib(n))
 
# This code is contributed
# by ChitraNayal

C＃

// Fibonacci Series using
// Optimized Method
using System;
 
class GFG
{
/* function that returns
nth Fibonacci number */
static int fib(int n)
{
int[,] F = new int[,]{{1, 1},
                      {1, 0}};
if (n == 0)
    return 0;
power(F, n - 1);
 
return F[0, 0];
}
 
static void multiply(int[,] F,
                     int[,] M)
{
int x = F[0, 0] * M[0, 0] +
        F[0, 1] * M[1, 0];
int y = F[0, 0] * M[0, 1] +
        F[0, 1] * M[1, 1];
int z = F[1, 0] * M[0, 0] +
        F[1, 1] * M[1, 0];
int w = F[1, 0] * M[0, 1] +
        F[1, 1] * M[1, 1];
 
F[0, 0] = x;
F[0, 1] = y;
F[1, 0] = z;
F[1, 1] = w;
}
 
/* Optimized version of
power() in method 4 */
static void power(int[,] F, int n)
{
if( n == 0 || n == 1)
return;
int[,] M = new int[,]{{1, 1},
                      {1, 0}};
 
power(F, n / 2);
multiply(F, F);
 
if (n % 2 != 0)
multiply(F, M);
}
 
// Driver Code
public static void Main ()
{
    int n = 9;
    Console.Write(fib(n));
}
}
 
// This code is contributed
// by ChitraNayal

Javascript

 
// Fibonacci Series using Optimized Method
 
// Function that returns nth Fibonacci number
function fib(n)
{
    var F = [ [ 1, 1 ], [ 1, 0 ] ];
    if (n == 0)
        return 0;
         
    power(F, n - 1);
 
    return F[0][0];
}
 
function multiply(F, M)
{
    var x = F[0][0] * M[0][0] + F[0][1] * M[1][0];
    var y = F[0][0] * M[0][1] + F[0][1] * M[1][1];
    var z = F[1][0] * M[0][0] + F[1][1] * M[1][0];
    var w = F[1][0] * M[0][1] + F[1][1] * M[1][1];
 
    F[0][0] = x;
    F[0][1] = y;
    F[1][0] = z;
    F[1][1] = w;
}
 
// Optimized version of power() in method 4 */
function power(F, n)
{
    if (n == 0 || n == 1)
        return;
         
    var M = [ [ 1, 1 ], [ 1, 0 ] ];
 
    power(F, n / 2);
    multiply(F, F);
 
    if (n % 2 != 0)
        multiply(F, M);
}
 
// Driver code
var n = 9;
 
document.write(fib(n));
 
// This code is contributed by gauravrajput1
 

输出

34

时间复杂度： O(Logn)
额外空间：如果我们考虑函数调用堆栈大小，则为 O(Logn)，否则为 O(1)。

方法 6(O(Log n) 时间)
下面是一个更有趣的递归公式，可用于在 O(Log n) 时间内找到第 n 个斐波那契数。

If n is even then k = n/2:
F(n) = [2*F(k-1) + F(k)]*F(k)

If n is odd then k = (n + 1)/2
F(n) = F(k)*F(k) + F(k-1)*F(k-1)

这个公式是如何工作的?
该公式可以从上述矩阵方程推导出来。

Taking determinant on both sides, we get 

(-1)n = Fn+1Fn-1 - Fn2 
 
Moreover, since AnAm = An+m for any square matrix A, 
the following identities can be derived (they are obtained 
form two different coefficients of the matrix product)

FmFn + Fm-1Fn-1 = Fm+n-1         ---------------------------(1)

By putting n = n+1 in equation(1),
FmFn+1 + Fm-1Fn = Fm+n             --------------------------(2)

Putting m = n in equation(1).
F2n-1 = Fn2 + Fn-12
Putting m = n in equation(2)

F2n = (Fn-1 + Fn+1)Fn = (2Fn-1 + Fn)Fn (Source: Wiki)   --------
( By putting Fn+1 = Fn + Fn-1 )
To get the formula to be proved, we simply need to do the following 
If n is even, we can put k = n/2 
If n is odd, we can put k = (n+1)/2

下面是上述想法的实现。

C++

// C++ Program to find n'th fibonacci Number in
// with O(Log n) arithmetic operations
#include 
using namespace std;
 
const int MAX = 1000;
 
// Create an array for memoization
int f[MAX] = {0};
 
// Returns n'th fibonacci number using table f[]
int fib(int n)
{
    // Base cases
    if (n == 0)
        return 0;
    if (n == 1 || n == 2)
        return (f[n] = 1);
 
    // If fib(n) is already computed
    if (f[n])
        return f[n];
 
    int k = (n & 1)? (n+1)/2 : n/2;
 
    // Applying above formula [Note value n&1 is 1
    // if n is odd, else 0.
    f[n] = (n & 1)? (fib(k)*fib(k) + fib(k-1)*fib(k-1))
           : (2*fib(k-1) + fib(k))*fib(k);
 
    return f[n];
}
 
/* Driver program to test above function */
int main()
{
    int n = 9;
    printf("%d ", fib(n));
    return 0;
}

Java

// Java Program to find n'th fibonacci
// Number with O(Log n) arithmetic operations
import java.util.*;
 
class GFG {
     
    static int MAX = 1000;
    static int f[];
     
    // Returns n'th fibonacci number using
    // table f[]
    public static int fib(int n)
    {
        // Base cases
        if (n == 0)
            return 0;
             
        if (n == 1 || n == 2)
            return (f[n] = 1);
      
        // If fib(n) is already computed
        if (f[n] != 0)
            return f[n];
      
        int k = (n & 1) == 1? (n + 1) / 2
                            : n / 2;
      
        // Applying above formula [Note value
        // n&1 is 1 if n is odd, else 0.
        f[n] = (n & 1) == 1? (fib(k) * fib(k) +
                        fib(k - 1) * fib(k - 1))
                       : (2 * fib(k - 1) + fib(k))
                       * fib(k);
      
        return f[n];
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int n = 9;
        f= new int[MAX];
        System.out.println(fib(n));
    }
}
     
// This code is contributed by Arnav Kr. Mandal.

Python

# Python3 Program to find n'th fibonacci Number in
# with O(Log n) arithmetic operations
MAX = 1000
 
# Create an array for memoization
f = [0] * MAX
 
# Returns n'th fibonacci number using table f[]
def fib(n) :
    # Base cases
    if (n == 0) :
        return 0
    if (n == 1 or n == 2) :
        f[n] = 1
        return (f[n])
 
    # If fib(n) is already computed
    if (f[n]) :
        return f[n]
 
    if( n & 1) :
        k = (n + 1) // 2
    else :
        k = n // 2
 
    # Applying above formula [Note value n&1 is 1
    # if n is odd, else 0.
    if((n & 1) ) :
        f[n] = (fib(k) * fib(k) + fib(k-1) * fib(k-1))
    else :
        f[n] = (2*fib(k-1) + fib(k))*fib(k)
 
    return f[n]
 
 
# Driver code
n = 9
print(fib(n))
 
 
# This code is contributed by Nikita Tiwari.

C＃

// C# Program to find n'th
// fibonacci Number with
// O(Log n) arithmetic operations
using System;
 
class GFG
{
 
static int MAX = 1000;
static int[] f;
 
// Returns n'th fibonacci
// number using table f[]
public static int fib(int n)
{
    // Base cases
    if (n == 0)
        return 0;
         
    if (n == 1 || n == 2)
        return (f[n] = 1);
 
    // If fib(n) is already
    // computed
    if (f[n] != 0)
        return f[n];
 
    int k = (n & 1) == 1 ? (n + 1) / 2
                         : n / 2;
 
    // Applying above formula
    // [Note value n&1 is 1 if
    // n is odd, else 0.
    f[n] = (n & 1) == 1 ? (fib(k) * fib(k) +
                           fib(k - 1) * fib(k - 1))
                        : (2 * fib(k - 1) + fib(k)) *
                                            fib(k);
 
    return f[n];
}
 
// Driver Code
static void Main()
{
    int n = 9;
    f = new int[MAX];
    Console.WriteLine(fib(n));
}
}
 
// This code is contributed by mits

Javascript

      // JavaScript Program to find n'th fibonacci Number in
      // with O(Log n) arithmetic operations
 
      const MAX = 1000;
 
      // Create an array for memoization
      var f = [...Array(MAX)];
      f.fill(0);
 
      // Returns n'th fibonacci number using table f[]
      function fib(n) {
        // Base cases
        if (n == 0) return 0;
        if (n == 1 || n == 2) return (f[n] = 1);
 
        // If fib(n) is already computed
        if (f[n]) return f[n];
 
        var k = n & 1 ? (n + 1) / 2 : n / 2;
 
        // Applying above formula [Note value n&1 is 1
        // if n is odd, else 0.
        f[n] =
          n & 1
            ? fib(k) * fib(k) + fib(k - 1) * fib(k - 1)
            : (2 * fib(k - 1) + fib(k)) * fib(k);
 
        return f[n];
      }
 
      /* Driver program to test above function */
 
      var n = 9;
      document.write(fib(n));
       
      // This code is contributed by rdtank.
    

输出

34

该解决方案的时间复杂度为 O(Log n)，因为我们在每次递归调用中将问题一分为二。

方法七

另一种方法(使用公式)：
在这种方法中，我们直接实现了斐波那契数列中第 n 项的公式。
F _n = {[(√5 + 1)/2] ^ n} / √5

参考：http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.html

C++

// C++ Program to find n'th fibonacci Number
#include
#include
 
int fib(int n) {
  double phi = (1 + sqrt(5)) / 2;
  return round(pow(phi, n) / sqrt(5));
}
 
// Driver Code
int main ()
{
  int n = 9;
  std::cout << fib(n) << std::endl;
  return 0;
}
//This code is contributed by Lokesh Mohanty.

C

// C Program to find n'th fibonacci Number
#include
#include
int fib(int n) {
  double phi = (1 + sqrt(5)) / 2;
  return round(pow(phi, n) / sqrt(5));
}
int main ()
{
  int n = 9;
  printf("%d", fib(n));
  return 0;
}

Java

// Java Program to find n'th fibonacci Number
import java.util.*;
 
class GFG {
 
static int fib(int n) {
double phi = (1 + Math.sqrt(5)) / 2;
return (int) Math.round(Math.pow(phi, n)
                        / Math.sqrt(5));
}
 
// Driver Code
public static void main(String[] args) {
        int n = 9;
    System.out.println(fib(n));
    }
}
// This code is contributed by PrinciRaj1992

蟒蛇3

# Python3 program to find n'th
# fibonacci Number
import math
 
def fibo(n):
    phi = (1 + math.sqrt(5)) / 2
 
    return round(pow(phi, n) / math.sqrt(5))
     
# Driver code   
if __name__ == '__main__':
     
    n = 9
     
    print(fibo(n))
     
# This code is contributed by prasun_parate

C＃

// C# Program to find n'th fibonacci Number
using System;
 
public class GFG
{
    static int fib(int n)
    {
    double phi = (1 + Math.Sqrt(5)) / 2;
    return (int) Math.Round(Math.Pow(phi, n)
                            / Math.Sqrt(5));
    }
     
    // Driver code
    public static void Main()
    {
        int n = 9;
        Console.WriteLine(fib(n));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

    // Javascript Program to find n'th fibonacci Number
    function fib(n) {
      let phi = (1 + Math.sqrt(5)) / 2;
      return Math.round(Math.pow(phi, n) / Math.sqrt(5));
    }
     
    let n = 9;
      document.write(fib(n));
 
// This code is contributed by mukesh07.

输出

34

时间复杂度： O(logn)，这是因为计算phi^n需要logn时间
空间复杂度： O(1)

方法八

DP 使用记忆化(自上而下的方法)

我们可以通过存储到目前为止计算的斐波那契数来避免方法 1 中所做的重复工作。我们只需要将所有值存储在一个数组中。

C++

#include 
using namespace std;
int dp[10];
int fib(int n)
{
    if (n <= 1)
        return n;
 
    // temporary variables to store
    //  values of fib(n-1) & fib(n-2)
    int first, second;
 
    if (dp[n - 1] != -1)
        first = dp[n - 1];
    else
        first = fib(n - 1);
 
    if (dp[n - 2] != -1)
        second = dp[n - 2];
    else
        second = fib(n - 2);
 
    // memoization
    return dp[n] = first + second;
}
 
// Driver Code
int main()
{
    int n = 9;
 
    memset(dp, -1, sizeof(dp));
 
    cout << fib(n);
    getchar();
    return 0;
 
    // This code is contributed by Bhavneet Singh
}

Java

import java.util.*;
 
class GFG{
 
// Initialize array of dp
static int[] dp = new int[10];
 
static int fib(int n)
{
    if (n <= 1)
        return n;
         
    // Temporary variables to store
    // values of fib(n-1) & fib(n-2)
    int first, second;
     
    if (dp[n - 1] != -1)
        first = dp[n - 1];
    else
        first = fib(n - 1);
 
    if (dp[n - 2] != -1)
        second = dp[n - 2];
    else
        second = fib(n - 2);
 
    // Memoization
    return dp[n] = first + second;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 9;
 
    Arrays.fill(dp, -1);
 
    System.out.print(fib(n));
}
}
 
// This code is contributed by sujitmeshram

蟒蛇3

# Initialize array of dp
dp = [-1 for i in range(10)]
 
def fib(n):
    if (n <= 1):
        return n;
    global dp;
     
    # Temporary variables to store
    # values of fib(n-1) & fib(n-2)
    first = 0;
    second = 0;
 
    if (dp[n - 1] != -1):
        first = dp[n - 1];
    else:
        first = fib(n - 1);
    if (dp[n - 2] != -1):
        second = dp[n - 2];
    else:
        second = fib(n - 2);
    dp[n] = first + second;
 
    # Memoization
    return dp[n] ;
 
# Driver Code
if __name__ == '__main__':
    n = 9;
    print(fib(n));
 
# This code contributed by Rajput-Ji

C＃

using System;
class GFG {
     
    // Initialize array of dp
    static int[] dp = new int[10];
    static int fib(int n)
    {
        if (n <= 1)
            return n;
              
        // Temporary variables to store
        // values of fib(n-1) & fib(n-2)
        int first, second;
          
        if (dp[n - 1] != -1)
            first = dp[n - 1];
        else
            first = fib(n - 1);
      
        if (dp[n - 2] != -1)
            second = dp[n - 2];
        else
            second = fib(n - 2);
      
        // Memoization
        return dp[n] = first + second;
    }
 
  // Driver code
  static void Main()
  {
    int n = 9;
    Array.Fill(dp, -1);
    Console.Write(fib(n));
  }
}
 
// This code is contributed by divyeshrabadiya07.

Javascript

// Initialize array of dp
dp = Array.from({length: 10}, (_, i) => -1);
 
function fib(n)
{
    if (n <= 1)
        return n;
         
    // Temporary variables to store
    // values of fib(n-1) & fib(n-2)
    var first, second;
     
    if (dp[n - 1] != -1)
        first = dp[n - 1];
    else
        first = fib(n - 1);
 
    if (dp[n - 2] != -1)
        second = dp[n - 2];
    else
        second = fib(n - 2);
 
    // Memoization
    return dp[n] = first + second;
}
 
// Driver Code
var n = 9;
document.write(fib(n));
 
// This code is contributed by Amit Katiyar

输出

34

https://www.youtube.com/watch?v=LwZRsM7qhrI
该方法由 Chirag Agarwal 贡献。

参考：
http://en.wikipedia.org/wiki/Fibonacci_number
http://www.ics.uci.edu/~eppstein/161/960109.html

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