斐波那契数列三种实现（上台阶）-Python

# 递归
def fibs(n):
    if n < 1:
        return 0
    elif n == 1:
        return 1
    return fibs(n - 2) + fibs(n - 1)

print(fibs(6))

# 循环
def fibs(n):
    a = [1, 1]
    for i in range(n - 2):
        a.append(a[-2] + a[-1])
    return a[n-1]

print(fibs(6))

class Solution:
    def fib(self, n):
        if n==1 or n==2:
            return n
        a=1;b=2;c=3
        for i in range(3,n+1):
            c=a+b;a=b;b=c
        return c

class Solution:
    def climbStairs(self, n):
        def fact(n):
            result=1
            for i in range(1,n+1):
                result*=i
            return result
        total=0
        for i in range(n/2+1):
            total+=fact(i+n-2*i)/fact(i)/fact(n-2*i)
        return total