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 : 9 * 10 - 99 : 90 * 100-999 : 900 * ... */ public class Solution { public int findNthDigit(int n) { long count = 9; // 因为要 9 ＊ 1000000000 注意会超出最大整数 ！！！！ int len = 1; int start = 1; while(n > count * len){ n -= count * len; count *= 10; len += 1; start *= 10; } start += (n -1) / len; String c = Integer.toString(start); return Character.getNumericValue(c.charAt((n-1) % len)); } }
原文：http://www.cnblogs.com/joannacode/p/6130567.html