10：斐波那切数列

大家都知道斐波那契数列，现在要求输入一个整数n，请你输出斐波那契数列的第n项。n<=39
n=0时，f(n)=0 n=1时，f(n)=1 n>1时，f(n)=f(n-1)+f(n-2)
法一：用循环较快

class Solution:
    def fib(self, x):
        if x<=0:
            return 0
        elif x ==1:
            return 1
        else:
            n, a, b = 0, 0, 1
            while n+1 != x:
                a, b = b, a + b
                n += 1
            return b
a = Solution()
b = a.fib(4)
print(b)

法二：用递归，较慢

class Solution:
    def fib(self, x):
        if x<= 0:
            return 0
        elif x==1:
            return 1
        else:
            return self.fib(x-1)+self.fib(x-2)
a = Solution()
b = a.fib(4)
print(b)

来源：https://blog.csdn.net/qq_43275748/article/details/102778053