边工作边刷题：70天一遍leetcode: day 61

Number of Digit One

要点：从低到高每一位累加，需要highbits，lowbits，base。基本的rule是每隔base会有1个one，一共多少个base？有3种情况>1, 1, <1。>1有highbits+1，<1有highbits，1有highbits还要加上lowbits+1
图
另一个重点是highbits，lowbits，curbit和base的关系：base是从1开始，和curbit是对应的。所以highbits是curbit左边的数，所以要/(base*10)。lowbits是右边的数，所以%base（所以在个位为%1==0）。curbit就是/base%10
错误点：

• loop invariant是n/base>0，而不是n
• highbits/lowbits/curbit要在loop内更新

class Solution(object):
    def countDigitOne(self, n):
        """
        :type n: int
        :rtype: int
        """
        base = 1
        count = 0
        while n/base>0:
            highbits = n//(base*10)
            lowbits = n%base
            curbit = n%(base*10)//base
            if curbit<1:
                count+=highbits*base
            elif curbit==1:
                count+=highbits*base+lowbits+1
            elif curbit>1:
                count+=(highbits+1)*base
            base*=10
        
        return count