求解斐波那契数列

斐波那契数列:0,1,1,2,3,5,8...

  1. 递归
int fib(int n){
    if(n == 0)
        return 0;
    else if(n == 1)
        return 1;
    else
        return fib(n-2) + fib(n-1); 
}
  1. 迭代
int fib(int n){
    if(n == 0)
        return 0;
    else if(n == 1)
        return 1;
    int num;
    int fib_1 = 0;
    int fib_2 = 1;
    for(int i=2; i<=n; i++){
        num = fib_1 + fib_2;
        fib_1 = fib_2;
        fib_2 = num;
    }
    return num;
}

你可能感兴趣的:(求解斐波那契数列)