400. Nth Digit

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))); } }}

版权声明：本文内容由网络用户投稿，版权归原作者所有，本站不拥有其著作权，亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容，请联系我们jiasou666@gmail.com 处理，核实后本网站将在24小时内删除侵权内容。