递归求1-i的平方和
Python 3.6.1 (default, Mar 22 2017, 06:17:05)
[GCC 6.3.0 20170321] on linux
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>>> class A:
... def func(self, i):
... return i * i
... def digui(self, i):
... if i == 1:
... return 1
... return self.func(i) + self.digui(i-1)
...
>>> A().digui(3)
14
>>> A().digui(2)
5
>>> A().digui(1)
1
完全可以啊,和函数一样的写法。只是多一个参数self
>>> class recursion:
... def method(self,n):
... if n==1:return 1
... return n*self.method(n-1)
...
>>> a=recursion()
>>> a.method(5)
120
就像普通递归一样调用