关于斐波那契数列：如何优化时间复杂度

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1.原始写法

 

def fib(n):
    if n<=0:
        return 0
    elif n == 1:
        return 1
    else:
        Return fib(n-1)+fib(n-2)

 

2.优化写法：定义一个字典记录已经计算过的

Know = {0:0,1:1}
def fib(n):
    if n in Know:
        Return Know[n]
    else:
        Res = fib(n-1)+fib(n-2)
        Know[n]=res
        return res

参考资料

https://blog.csdn.net/feiyu07/article/details/73250558