About Python maximum recursion depth-998

Today, Leetcode, the Brute force solution 233

• Problem:

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

For example:

Given n = 13,

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

• Code:

classSolution:def __init__( Self): Self. Key= ' 1 '         Self. Result= 0    defCountdigitone ( Self, N):""": Type N:int: Rtype:int        """        ifN< 1:return  Self. Result Self. Result+= Str(n). Count ( Self. Key)ifN> 0: Self. Countdigitone (n-1)return  Self. Results=Solution ()Print(S.countdigitone (11221))

• Error:

Maximum recursion depth exceeded while getting the STR in an object

• Find Python maximum recursion depth

classSolution:def __init__( Self): Self. Key= ' 1 '         Self. Result= 0    defCountdigitone ( Self, N):""": Type N:int: Rtype:int        """        ifN< 1:return  Self. Result Self. Result+= Str(n). Count ( Self. Key)ifN> 0: Self. Countdigitone (n-1)return  Self. Results=Solution () forIinch Range(0,1000000):Print(i)Print(S.countdigitone (i))

Output 998, then error, maximum recursion depth found, or peace of mind with while bar ~

About Python maximum recursion depth-998