剑指offer-斐波那契数列-递归和循环-python

# -*- coding:utf-8 -*-
class Solution:
    def Fibonacci(self, n):
        # write code here
        #递归法
        if n ==0:
            return 1
        elif n ==1:
            return 1
        else:
            return self.Fibonacci(n-1) +self.Fibonacci(n-2)
            

# -*- coding:utf-8 -*-
class Solution:
    def Fibonacci(self, n):
        # write code here
        #递归法
        a = 0
        b = 1
        # 循环叠加，因为每次叠加两个，所以，只需要循环到n的一半就行，但是数据是从0开始，所以需要到n // 2 + 1
        for i in range(1, n // 2 + 1):
            a = a + b
            b = a + b
        # 叠加之后，如果n为奇数，那么返回b，如果n为偶数，那么返回a
        if n % 2 == 1:
            return b
        else:
            return a
        pass

class Solution:
    def Fibonacci(self, n):
        # write code here
        a = 0
        b = 1
        if n <= 1:
            return n
        for i in range(n):
            a ,b = b,b+a            
        return a