标签：return sequence ret logs ber find input bsp init
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 class Solution { 2 public: 3 int findNthDigit(int n) { 4 // step 1. calculate how many digits the number has. 5 long base = 9, digits = 1; 6 while (n - base * digits > 0) 7 { 8 n -= base * digits; 9 base *= 10; 10 digits ++; 11 } 12 13 // step 2. calculate what the number is. 14 int index = n % digits; 15 if (index == 0) 16 index = digits; 17 long num = 1; 18 for (int i = 1; i < digits; i ++) 19 num *= 10; 20 num += (index == digits) ? n / digits - 1 : n / digits;; 21 22 // step 3. find out which digit in the number is we wanted. 23 for (int i = index; i < digits; i ++) 24 num /= 10; 25 return num % 10; 26 } 27 };
标签：return sequence ret logs ber find input bsp init
原文地址：http://www.cnblogs.com/wujufengyun/p/7197209.html
