计算数字位数而无需字符串操作

Question

Im trying to write a function which counts the number of digits in a number - with no string operation. here is my code:

def count_digits(n):
    count_list=[]
    while (n>0):
        n=n%10
        i=n
        count_list.append(i)
        n=n/10
        return len(count_list)

n=12345
print count_digits(n)

By using % i get the last digits - in order to add it to a list. By using / i throw the digit from the number.

The script does not work. For each n i put, the script just prints 1 .

Thanks!

Answer 1

There are several problems to your code:

• The return statement should be outside of the loop.
• The n = n % 10 statement modifies n , making it impossible to get the other digits.
• I'd use the integer division operator // . In Python 3, n / 10 would give a floating point number.
• Like ShadowRanger said, the current solution considers 0 as having 0 digits. You need to check whether n is 0 .

Here is a corrected version of your code:

def count_digits(n):
    if n == 0:
        return 1
    count_list = []
    while n > 0:
        count_list.append(n % 10)
        n = n // 10
    return len(count_list)

Also, as it was said in the comments, since your goal is just to count the digits, you don't need to maintain a list:

def count_digits(n):
    if n == 0:
        return 1
    count = 0
    while n > 0:
        n = n // 10
        count += 1
    return count

Answer 2

count_list stores the digits.

def count_digits(n):
    count_list=[]
    while (n>0):
        count_list.append(n%10)
        n-=n%10
        n = n/10
    return count_list

n=12345
print len(count_digits(n))

Without using a list

def count_digits(n):
   count_list=0
   while (n>0):
       count_list+=1
       n-=n%10
       n = n/10
   return count_list

n=12345
print count_digits(n)

Answer 3

Perhaps you can try this, a much simpler approach. No lists involved :)

def dcount(num):
    count = 0
    if num == 0:
        return 1
    while (num != 0):
        num /= 10
        count += 1
    return count