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)