算法题2——斐波那契数列
计算斐波那契数列的第 n 个数:
def fibonacci(n):
if n <= 0:
return 0
if n == 1:
return 1
a, b = 0, 1
for _ in range(2, n + 1):
a, b = b, a + b
return b
计算斐波那契数列的第 n 个数:
def fibonacci(n):
if n <= 0:
return 0
if n == 1:
return 1
a, b = 0, 1
for _ in range(2, n + 1):
a, b = b, a + b
return b