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.
class Solution { public: int findNthDigit(int n) { // 1. calc how many digits the number has. long base = 9, digits = 1; while (n - base * digits > 0) { n -= base * digits; base *= 10; digits++; } // 2. calc what the number is. int index = n % digits; if (index == 0) index = digits; long num = 1; for (int i = 1; i < digits; i++) num *= 10; num += (index == digits) ? n / digits - 1 : n / digits; // 3. find out which digit in the number. for (int i = index; i < digits; i++) num /= 10; return num % 10; } }; // 3 ms
原文：http://www.cnblogs.com/immjc/p/7994218.html