"Write a recursive function, "listSum" that takes a list of integers and returns the sum of all integers in the list".

>>>> listSum([1,3,4,5,6])
19

def listSum(ls):
    i = 0
    s = 0
    while i < len(ls):
        s = s + ls[i]
        i = i + 1
    print s

最佳答案

每当你遇到这样的问题时，试着用相同的函数来表达函数的结果。

在您的情况下，您可以通过将第一个数字与对列表中的其余元素调用相同函数的结果相加来获得结果。

例如，

listSum([1, 3, 4, 5, 6]) = 1 + listSum([3, 4, 5, 6])
                         = 1 + (3 + listSum([4, 5, 6]))
                         = 1 + (3 + (4 + listSum([5, 6])))
                         = 1 + (3 + (4 + (5 + listSum([6]))))
                         = 1 + (3 + (4 + (5 + (6 + listSum([])))))

>>> def listSum(ls):
...     # Base condition
...     if not ls:
...         return 0
...
...     # First element + result of calling `listsum` with rest of the elements
...     return ls[0] + listSum(ls[1:])
>>> 
>>> listSum([1, 3, 4, 5, 6])
19

>>> def listSum(ls, result):
...     if not ls:
...         return result
...     return listSum(ls[1:], result + ls[0])
... 
>>> listSum([1, 3, 4, 5, 6], 0)
19

>>> def listSum(ls, index, result):
...     # Base condition
...     if index == len(ls):
...         return result
...
...     # Call with next index and add the current element to result
...     return listSum(ls, index + 1, result + ls[index])
... 
>>> listSum([1, 3, 4, 5, 6], 0, 0)
19

>>> def listSum(ls):
...
...     def recursion(index, result):
...         if index == len(ls):
...             return result
...         return recursion(index + 1, result + ls[index])
...
...     return recursion(0, 0)
... 
>>> listSum([1, 3, 4, 5, 6])
19

>>> def listSum(ls, index=0, result=0):
...     # Base condition
...     if index == len(ls):
...         return result
...
...     # Call with next index and add the current element to result
...     return listSum(ls, index + 1, result + ls[index])
... 
>>> listSum([1, 3, 4, 5, 6])
19

power(2, 5) = 32
power(5, 2) = 25
power(3, 4) = 81

power(2, 5) = 2 * 2 * 2 * 2 * 2 = 32
power(5, 2) = 5 * 5             = 25
power(3, 4) = 3 * 3 * 3 * 3     = 81

power(2, 5) = 2 * power(2, 4)
            = 2 * (2 * power(2, 3))
            = 2 * (2 * (2 * power(2, 2)))
            = 2 * (2 * (2 * (2 * power(2, 1))))

            = 2 * (2 * (2 * (2 * 2)))
            = 2 * (2 * (2 * 4))
            = 2 * (2 * 8)
            = 2 * 16
            = 32

>>> def power(base, exponent):
...     # Base condition, if `exponent` is lesser than or equal to 1, return `base`
...     if exponent <= 1:
...         return base
...
...     return base * power(base, exponent - 1)
... 
>>> power(2, 5)
32
>>> power(5, 2)
25
>>> power(3, 4)
81

>>> def power(base, exponent, result=1):
...     # Since we start with `1`, base condition would be exponent reaching 0
...     if exponent <= 0:
...         return result
...
...     return power(base, exponent - 1, result * base)
... 
>>> power(2, 5)
32
>>> power(5, 2)
25
>>> power(3, 4)
81

power(2, 5, 1) = power(2, 4, 1 * 2)
               = power(2, 4, 2)
               = power(2, 3, 2 * 2)
               = power(2, 3, 4)
               = power(2, 2, 4 * 2)
               = power(2, 2, 8)
               = power(2, 1, 8 * 2)
               = power(2, 1, 16)
               = power(2, 0, 16 * 2)
               = power(2, 0, 32)

关于python - Python 中的递归基础，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/30214531/