400. 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: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.

思路：
1~9 9个数 9*1=9个digit
10~99 90个数 90*2=180个digit
100~999 900个数 900*3=2700个digit
10^k ~ k个9连成的数 9*10^k个数 (90*10^k)*k个digit

class Solution {    public int findNthDigit(int n) {        int k = 1;        long len = 0;        while (n > len) {            len += (int) (9 * k * Math.pow(10, k - 1));            k++;        }        k--;        len -= (int) (9 * k * Math.pow(10, k - 1));        int num = 0;        if ((n - len) % k == 0) {            num = (int) ((int) Math.pow(10, k - 1) + (n - len) / k - 1);            String s = String.valueOf(num);            return Integer.parseInt("" + s.charAt(s.length() - 1));        } else {            num = (int) ((int) Math.pow(10, k - 1) + (n - len) / k);            String s = String.valueOf(num);            return Integer.parseInt("" + s.charAt((int) ((n - len) % k - 1)));        }    }}