Leetcode: Nth Digit

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.

1-9 : count:9 * len:1

10-99: count:90 * len:2

100-999: count:900 * len:3

1000-9999: count: 9000 * len:4

maintain a count, len, start

 1 public class Solution {
 2     public int findNthDigit(int n) {
 3         int start = 1;
 4         int len = 1;
 5         long count = 9;
 6         while (n > len*count) {
 7             n -= len*count;
 8             start *= 10;
 9             len ++;
10             count *= 10;
11         }
12         start += (n-1)/len;
13         char res = Integer.toString(start).charAt((n-1)%len);
14         return Character.getNumericValue(res);
15     }
16 }
原文地址:https://www.cnblogs.com/EdwardLiu/p/6100243.html