算法：求从1到n这n个整数的十进制表示中1出现的次数 python 实现

题目：输入一个整数n，求从1到n这n个整数的十进制表示中1出现的次数。
例如输入12，从1到12这些整数中包含1 的数字有1，10，11和12，1一共出现了5次。
据说这是一道google面试题,在何海涛的博客(http://zhedahht.blog.163.com/blog/static/25411174200732494452636/)中已有递归解法，
在阳振坤的博客中提出了一种非递归的算法（http://blog.sina.com.cn/s/blog_3fc85e260100mbss.html)
在这里，我给出这个算法的python实现:

 

$cat countone.py
#!/usr/bin/python
#
# Find the number of 1 in the integers between 1 and N
# Input: N - an integer
# Output: the number of 1 in the integers between 1 and N
#
def CountOne(N):
    #make sure that N is an integer
    N = int(N)
    #convert N to chars
    a = str(N)
    #the lenth is string a
    n = len(a)
    i = 0
    count = 0
    while (i < n):
       if(i == 0):
           if(int(a[i]) == 1 ):
               count += int(a[1:])+1
           elif(int(a[i]) > 1):
               count += 10 ** (n-1)
       elif(i == n - 1):
           if(int(a[i]) == 0):
               count += int(a[:n-1])
           else:
               count += int(a[:n-1]) + 1
       else:
           if(int(a[i]) == 0):
               count += int(a[:i]) * (10 ** (n - i - 1))
           elif(int(a[j]) == 1):
               count += int(a[:i]) * (10 ** (n - i - 1)) + int(a[i+1:]) + 1
           else:
               count += (int(a[:i]) + 1) * (10 ** (n - i -1))
       i += 1
    return count

#Test code
import sys
if(__name__ == "__main__"):
    N = int(sys.argv[1])
    n = CountOne(N)
    print "the number of '1' between 1 and %d is %d" % (N,n)