如何使用递归查找加到 10 的数字

Question

I'm trying to write a function that takes an int as a parameter and finds the number of pairs of numbers within that add to 10.

For example, 7645238 has 3 pairs because: 7 + 3 = 10, 6 + 4 = 10, and 2 + 8 = 10.

I have to do this recursively and I have most of the code done, I just have a small problem. I'm using a loop and counter variable in my recursive function, but the counter variable resets each time the function loops. Thus, if a number has more than one pair that adds to 10, the counter will always only return 1.

Here's my code in python:

def findPairs(num):

    count = 0
    num = str(num)

    # base case if length of num is 1
    if len(num) == 1:
        return 0
    # loops through the rest of the number looking for pairs
    else:
        for n in num[1:]:
            if int(num[0]) + int(n) == 10:
                count = count + 1
        findPairs(num[1:])
    return count

Any help would be greatly appreciated!

Answer 1

I think the problem is in the returning code, you should return the counter for the current integer + the recursive counters sum, so probably this code will fix the problem:

def findPairs(num):

count = 0
num = str(num)

# base case if length of num is 1
if len(num) == 1:
    return 0
# loops through the rest of the number looking for pairs
for n in num[1:]:
     if int(num[0]) + int(n) == 10:
         count = count + 1
return count + findPairs(num[1:])