Number of Digit One

题目：

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.

解题思路：

(1)暴力破解（O(n)）-

Time Limit Exceeded

class Solution:

# @param {integer} n

# @return {integer}

def countDigitOne(self, n):

res = 0

for i in range(1,n+1):

s = str(i)

res = res+s.count('1')

return res

(2)逐位求法

还有一种方法就是数每一个数位上1出现的个数。

对于一个数 abcd

当 d > 1的时候 个位上出现1的个数为 (abc + 1)*1

当d = 1 的时候 个位上出现1的个数为 abc * 1 + 0 + 1， 0表示 d后面的数。对于11 来说，个位上出现1的数为 11,1

当d = 0的时候 个位上出现1的个数为 abc * 1

对于其它数位，相对的1变为10,100,1000..
class Solution:

# @param {integer} n

# @return {integer}

def countDigitOne(self, n):

res = 0

base = 1

before_dig = 0

while n>0:

end_dig = n%10

n = n/10

if end_dig>1:

res = res+(n+1)*base

elif end_dig==1:

res = res+n*base+before_dig+1

else:

res = res+n*base

before_dig = end_dig*base+before_dig

base = base*10

return res