400. Nth Digit（LeetCode）

Posted 2020-09-28 无惧风云

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.

 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 };