Fibonacci斐波那契数列，也叫兔子数列，从第3项开始，每一项是前2项之和。递归定义如下：
F i b ( n ) = { 0 , n = 0 1 , n = 1 F i b ( n − 1 ) + F i b ( n − 2 ) , n > = 2 Fib(n)=\left\{\begin{array}{l} \qquad \qquad 0\qquad \qquad \qquad ,n=0 \\ \qquad \qquad 1\qquad \qquad \qquad , n=1 \\ Fib(n-1)+Fib(n-2), n>=2 \end{array}\right. Fib(n)=⎩⎨⎧​0,n=01,n=1Fib(n−1)+Fib(n−2),n>=2​

    Fibonacci斐波那契数列的前几项，依次为：0,1,1,2,3,5,8,21,34,…

    斐波那契数列，有通项公式，如下：

F i b ( n ) = [ ( 1 + 5 2 ) n − ( 1 − 5 2 ) n ] / 5 \mathrm{Fib}(n)=\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right] / \sqrt{5} Fib(n)=[(21+5 ​​)n−(21−5 ​​)n]/5 ​

    斐波那契数列的计算方法，有很多种，每种方法的时间复杂度、空间复杂度相差较大，具体如下：

1、斐波那契数列 C++实现

//fib.cpp

#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;

//普通递归实现
long Fib(long n) {
	if (n < 2)
		return n;
	else
		return Fib(n - 1) + Fib(n - 2);
}

//尾递归实现
long Fib2(long first,long second,long n) {
	if (n == 0)
		return 0;
	if (n == 1 || n == 2)
		return 1;
	if (n == 3)
		return first + second;
	return Fib2(second, first + second, n - 1);
}

//循环实现
long Fib3(long n) {
	if (n == 0)
		return 0;

	long first = 1, second = 1;
	long ret = 0;
	for (long i = 3; i <= n; i++){
		ret = first + second;
		first = second;
		second = ret;
	}
	return second;
}

#define Half_Sqrt5   1.1180339887498948482045868343656 
#define All_Sqrt5    2.2360679774997896964091736687313

//用通项公式实现
long Fib4(long n) {
	if (n < 2) {
		return n;
	}
	return (long)(round((pow(0.5+Half_Sqrt5,n) - pow(0.5-Half_Sqrt5,n))/All_Sqrt5));
}

int main(){
	
	//---- fib(6) = 8
	int Fib6a = Fib(6);
	cout << "Fib6a= " << Fib6a << endl;

	int Fib6b = Fib2(1,1,6);
	cout << "Fib6b= " << Fib6b << endl;

	int Fib6c = Fib3(6);
	cout << "Fib6c= " << Fib6c << endl;

	int Fib6d = Fib4(6);
	cout << "Fib6d= " << Fib6d << endl;
	

	system("pause");
	return 0;
}

    效果如下：

2、斐波那契数列 go实现

//fib_test.go

package yufa

import (
	"fmt"
	"math"
	"testing"
)

//普通递归实现
func Fib(n int64) (res int64) {
	if n < 2 {
		return n
	} else {
		return Fib(n-1)+Fib(n-2)
	}
}

//尾递归实现
func Fib2(first,second,n int64)(res int64){
	if n == 0 {
		return 0
	} else if n==1 || n==2 {
		return 1
	} else if n== 3 {
		return first+second
	}
	return Fib2(second, first+second, n-1)
}

//循环实现
func Fib3(n int64) (res int64) {
	if n == 0 {
		return 0
	}

	var first,second int64 = 1,1
	var ret,i int64 = 0,0
	for i=3; i<=n;i++ {
		ret = first + second
		first = second
		second = ret
	}
	return second
}

const (
	Half_Sqrt5 = 1.1180339887498948482045868343656
	All_Sqrt5  = 2.2360679774997896964091736687313
)

//四舍五入
func round(x float64) (res int64) {
	return int64(math.Floor(x+0.5))
}

//用通项公式实现
func Fib4(n int64) (res int64) {
	if n <= 1 {
		return n
	}

	res = round((math.Pow(0.5+Half_Sqrt5,float64(n)) - math.Pow(0.5-Half_Sqrt5,float64(n)))/All_Sqrt5)
	return res
}

func TestFib(t *testing.T) {
	var Fib6a int64 = Fib(6)
	var Fib6b int64 = Fib2(1,1,6)
	var Fib6c int64 = Fib3(6)
	var Fib6d int64 = Fib4(6)
	fmt.Println("Fib6a= ",Fib6a)
	fmt.Println("Fib6b= ",Fib6b)
	fmt.Println("Fib6c= ",Fib6c)
	fmt.Println("Fib6d= ",Fib6d)

}

    效果如下：

参考文献

[1] 力扣.斐波那契数列的6种解法.2019
[2] latex公式编辑