如果给定一个整数列表和一个名为 x 的数字，如何递归返回列表中每个第 x 个数字的总和

Question

So, How to if given a list of integers and a number called x, return recursively the sum of every x'th number in the list.

In this task "indexing" starts from 1, so if x = 2 and nums = [2, 3, 4, -9] , the output should be -6 (3 + -9).

X can also be negative, in that case indexing starts from the end of the list, see examples below.

If x = 0 , the sum should be 0 as well.

For example:

print(x_sum_recursion([], 3))  # 0
print(x_sum_recursion([2, 5, 6, 0, 15, 5], 3))  # 11
print(x_sum_recursion([0, 5, 6, -5, -9, 3], 1))  # 0
print(x_sum_recursion([43, 90, 115, 500], -2))  # 158
print(x_sum_recursion([1, 2], -9))  # 0
print(x_sum_recursion([2, 3, 6], 5))  # 0
print(x_sum_recursion([6, 5, 3, 2, 9, 8, 6, 5, 4], 3))  # 15

I have been trying to do this function for 5 hours straight!!!

Would like do see how someone else would solve this.

This is the best i have come up with.

def x_sum_rec_Four(nums: list, x: int) -> int:
    if len(nums) == 0:
        return 0
    elif len(nums) < x:
        return 0
    elif x > 0:
        i = x - 1
        return nums[i] + x_sum_rec_Four(nums[i + x:], x)
    elif x < 0:
        return x_sum_rec_Four(nums[::-1], abs(x))

My problem with this recursion is that the finish return should be:

if len(nums) < x:
    return nums[0]

but that would pass things like ([2, 3, 6], 5)) -->> 2 when it should be 0.

Answer 1

If you really need to do it recursively, you could pop x-1 elements from the list before each call, follow the comments below:

def x_sum_recursion(nums, x):
    # if x is negative, call the function with positive x and reversed list
    if x < 0:
        return x_sum_recursion(nums[::-1], abs(x))
    # base case for when x is greater than the length of the list
    if x > len(nums):
        return 0
    # otherwise remove the first x-1 items
    nums = nums[x-1:]
    # sum the first element and remove it from the next call
    return nums[0] + x_sum_recursion(nums[1:], x)

print(x_sum_recursion([], 3))  # 0
print(x_sum_recursion([2, 5, 6, 0, 15, 5], 3))  # 11
print(x_sum_recursion([0, 5, 6, -5, -9, 3], 1))  # 0
print(x_sum_recursion([43, 90, 115, 500], -2))  # 158
print(x_sum_recursion([1, 2], -9))  # 0
print(x_sum_recursion([2, 3, 6], 5))  # 0
print(x_sum_recursion([6, 5, 3, 2, 9, 8, 6, 5, 4], 3))  # 15

However, you can do it in a one liner simple and pythonic way:

print(sum(nums[x-1::x] if x > 0 else nums[x::x]))

Explanation:

you slice the list using nums[start:end:increment] when you leave the end empty it slices from the starting position till the end of the list and increments by the increment specified