[leetcode: Python]400. Nth Digit

题目：

Find the nth digit of the infinite integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, …

Note:

n is positive and will fit within the range of a 32-bit signed integer (n < 231).

Example 1:

Input:
3

Output:
3

Example 2:

Input:
11

Output:
0

Explanation:
The 11th digit of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... is a 0, which is part of the number 10

注意：

n是正数并且范围在32位带符号整数之内（n < 2^31）

解题思路：

将整数序列划分为下列区间：

1   1-9
2   10-99
3   100-999
4   1000-9999
5   10000-99999
6   100000-999999
7   1000000-9999999
8   10000000-99999999
9   100000000-99999999

然后分区间求值即可。

方法一：性能56ms

class Solution(object):
def findNthDigit(self, n):
"""
:type n: int
:rtype: int
"""
for i in range(9):
d = 9 * pow(10, i)
if n <= d * (i + 1): break
n -= d * (i+1)
n -= 1
return int(str(pow(10, i) + n/(i+1))[n % (i+1)])

方法二：性能38ms

class Solution(object):
def findNthDigit(self, n):
"""
:type n: int
:rtype: int
"""

N=1

while n>0:

r= 9* 10**(N-1)*N

if n>r:
n-=r
N+=1
else:
number= 10**(N-1) + (n-1)/N
return int(str(number)[(n-1)%N])