计算w/o将整数转换为字符串的位数:

x=123
x=abs(x)
i = 0
while x >= 10**i:
    i +=1
# i is the number of digits

这里是最简单的方法，不需要将int转换为字符串:

假设给出的数字为15位，例如;n = 787878899999999;

n=787878899999999 
n=abs(n) // we are finding absolute value because if the number is negative int to string conversion will produce wrong output

count=0 //we have taken a counter variable which will increment itself till the last digit

while(n):
    n=n//10   /*Here we are removing the last digit of a number...it will remove until 0 digits will left...and we know that while(0) is False*/
    count+=1  /*this counter variable simply increase its value by 1 after deleting a digit from the original number
print(count)   /*when the while loop will become False because n=0, we will simply print the value of counter variable

输入:

n=787878899999999

输出:

15

设数字为n，则n中的位数为:

math.floor(math.log10(n))+1

注意，这将为+ve个整数< 10e15给出正确答案。除此之外，返回类型的数学的精度限制。Log10开始起作用，结果可能相差1。我可以简单地在后面用len(str(n));这需要O(log(n))时间，相当于10的幂次迭代。

感谢@SetiVolkylany让我注意到这个限制。令人惊讶的是，看似正确的解决方案在实现细节中有警告。

def digits(n)
    count = 0
    if n == 0:
        return 1
    
    if n < 0:
        n *= -1

    while (n >= 10**count):
        count += 1
        n += n%10

    return count

print(digits(25))   # Should print 2
print(digits(144))  # Should print 3
print(digits(1000)) # Should print 4
print(digits(0))    # Should print 1

我的代码相同如下，我已经使用了log10方法:

from math import *

def digit_count(数量):

if number>1 and round(log10(number))>=log10(number) and number%10!=0 :
    return round(log10(number))
elif  number>1 and round(log10(number))<log10(number) and number%10!=0:
    return round(log10(number))+1
elif number%10==0 and number!=0:
    return int(log10(number)+1)
elif number==1 or number==0:
    return 1

我必须在1和0的情况下指定，因为log10(1)=0和log10(0)=ND，因此上面提到的条件不满足。但是，此代码仅适用于整数。

coin_digit = str(coin_fark).split(".")[1]
coin_digit_len = len(coin_digit)
print(coin_digit_len)