标签:python 面试题 algorithm
def Fibonacci(n): if n <= 0: return 0 if n <= 1: return n f0 = 0; f1 = 1 for i in range(2, n + 1): fn = f0 + f1 f0 = f1 f1 = fn return fn
【剑指offer】Q9:斐波那契数列,布布扣,bubuko.com
【剑指offer】Q9:斐波那契数列
原文地址:http://blog.csdn.net/shiquxinkong/article/details/37083045