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:3Output:3
Example 2:
Input:11Output:0Explanation: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.
给定一个无穷整数序列1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... 求序列的第n位数字。
1 2 3 4 5 6 7 8 9 10 11 12 | class Solution(object): def findNthDigit(self, n): """ :type n: int :rtype: int """ n -= 1 for digits in range(1, 11): first = 10**(digits - 1) if n < 9 * first * digits: return int(str(first + n/digits)[n%digits]) n -= 9 * first * digits |
原文：http://www.cnblogs.com/xiejunzhao/p/7203334.html