Python 返回递归函数的递归次数
全局变量
def fib(n):
global count
count += 1
return n if n <= 1 else fib(n-1) + fib(n-2)
count = 0
print(fib(5), count) # 5 15函数设置属性
def fib(n):
fib.count += 1
return n if n <= 1 else fib(n-1) + fib(n-2)
fib.count = 0
print(fib(5), fib.count) # 5类装饰器
class Counter(object) :
def __init__(self, fun) :
self._fun = fun
self.counter = 0
def __call__(self,*args, **kwargs) :
self.counter += 1
return self._fun(*args, **kwargs)
# 斐波那契数列
@Counter
def fib(n):
return n if n <= 1 else fib(n-1) + fib(n-2)
print(fib(5)) # 5
print(fib.counter) # 15
# 汉诺塔问题
@Counter
def move(n, a, b, c):
if n == 1:
print(a, '-->', c)
else:
move(n-1, a, c, b)
move(1, a, b, c)
move(n-1, b, a, c)
move(3, 'A', 'B', 'C')
print(move.counter) # 10
