233. Number of Digit One【H】【33】【再来一遍】

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本文介绍了一个算法问题：计算从0到n所有整数中数字1出现的总次数。通过逐位计算的方法，解决了该问题，并给出了Python实现代码。


Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.

For example:
Given n = 13,
Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.

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按位算，先算个位数的1 有多少个，再算十位数的，依次往上

写了好多遍啊。。。。 


class Solution(object):
    def countDigitOne(self, n):
        if n < 1:
            return 0

        b = 0
        t = 1
        res = 0

        while t <= n:
            b = n / (t*10)

            tres = b * t

            temp = (n / t)%10
            if temp == 10:
                temp = 1

            if temp == 1:
                tres += n % t + 1

            elif temp > 1:
                tres += t

            res += tres

            #print t,b,tres

            t *= 10

        return  int(res)


C++版 - 剑指offer 面试题32：从1到n整数中1出现的次数(leecode233. Number of Digit One) 题解
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剑指offer 面试题32：从1到n整数中1出现的次数(Leecode233. Number of Digit One) 提交网址： http://www.nowcoder.com/practice/bd7f978302044eee894445e244c7eee6?tpId=13&tqId=11184 题目： 输入一个整数n，求从1到n这n个整数的十进
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Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n. For example:Given n = 13,Return 6, because digit 1 occurred in the following numb...
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Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.Example:Input: 13 Output: 6 Explanation: Digit 1 occurred in the following numbers...
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