About python maximum recursion depth - 998

Violent solution 233 in LeetCode today

• question:

Given an integer n, count the number of occurrences of the number 1 in all non-negative numbers less than or equal to n.

E.g:

Given n = 13,

Returns 6, because the number 1 appears in the following numbers: 1, 10, 11, 12, 13.

• Code:

class Solution:
    def __init__(self):
        self.key = '1'
        self.result = 0

    def countDigitOne(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n < 1:
            return self.result
        self.result += str(n).count(self.key)
        if n > 0:
            self.countDigitOne(n-1)
        return self.result

s = Solution()
print(s.countDigitOne(11221))

• mistake:

maximum recursion depth exceeded while getting the str of an object

• Find python maximum recursion depth

class Solution:
    def __init__(self):
        self.key = '1'
        self.result = 0

    def countDigitOne(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n < 1:
            return self.result
        self.result += str(n).count(self.key)
        if n > 0:
            self.countDigitOne(n-1)
        return self.result

s = Solution()
for i in range(0,1000000):
    print(i)
    print(s.countDigitOne(i))

Output 998, and then report an error, the maximum recursion depth is found, or use while with peace of mind~