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.
对于这道题，要先确定输入的数字在几位数，通过规律可以知道，1位数的数字有9个，2位有90个，3位有900个，以此类推…………确定范围后，要找到该数字具体是哪一个，然后再确定是第几位，AC码如下，有一些思路的注释：
public class Solution { public int findNthDigit(int n) { int len = 1;//用于记录所在数字的位数 long count = 9; int start = 1; while (n > len * count) { n -= len * count; len += 1; count *= 10; start *= 10; } start += (n - 1) / len;//找到所在的那个数 String s = Integer.toString(start); return Character.getNumericValue(s.charAt((n - 1) % len)); }//确定具体在第几位 }