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.

class Solution {
public:
    int findNthDigit(int n) {
        
 
// 1. calc how many digits the number has.
        long base = 9, digits = 1;
        while (n - base * digits > 0) {
            n -= base * digits;
            base *= 10;
            digits++;
        }
        
        // 2. calc what the number is.
        int index = n % digits;
        if (index == 0
 
)
            index = digits;
        long num = 1;
        for (int i = 1; i < digits; i++)
            num *= 10;
        num += (index == digits) ? n / digits - 1 : n / digits;
        
        // 3. find out which digit in the number.
        for (int i = index; i < digits; i++)
            num /= 10;
        return num % 10;
    }
};
// 3 ms

[LeetCode] Nth Digit