题解 | #斐波那契数列#

F[0]=0,F[1]=1
F[n]=F[n-1]+Fn-2
法一：递归

function fibonacci(n) {
    //1.递归
   if(n==0) return 0
   else if(n==1) return 1
   else return fibonacci(n-1)+fibonacci(n-2)
    }

法二：循环

//2.循环
    let n1=0
    let n2=1
    let res
    if(n==0) return n1
    if(n==1) return n2
    while(n>=2){
       res=n1+n2;
        n1=n2;
        n2=res
        n--
    }
    return res

法二用解构赋值

     let n1=0
    let n2=1
    if(n==0) return 0
    if(n==1) return 1
    while(n>=2){
       [n1,n2]=[n2,n1+n2]
        n--
    }
    return n2