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.

1 class Solution { 2 public: 3     int findNthDigit(int n) { 4        // step 1. calculate how many digits the number has. 5         long base = 9, digits = 1; 6         while (n - base * digits > 0) 7         { 8             n -= base * digits; 9             base *= 10;10             digits ++;11         }12 13         // step 2. calculate what the number is.14         int index = n % digits;15         if (index == 0)16             index = digits;17         long num = 1;18         for (int i = 1; i < digits; i ++)19             num *= 10;20         num += (index == digits) ? n / digits - 1 : n / digits;;21 22         // step 3. find out which digit in the number is we wanted.23         for (int i = index; i < digits; i ++)24             num /= 10;25         return num % 10;26     }27 };