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.
求无穷序列里的第n个数字
比如第11个数是数字10里的0
C++（3ms）：
1 class Solution { 2 public: 3 int findNthDigit(int n) { 4 int len = 1 ; 5 long count = 9 ; 6 int start = 1 ; 7 while(n > len * count){ 8 n -= len * count ; 9 len+=1 ; 10 count *= 10 ; 11 start *= 10 ; 12 } 13 start += (n-1)/len ; 14 string s = to_string(start) ; 15 int res = s[(n-1)%len] - ‘0‘ ; 16 return res ; 17 } 18 };
java（7ms）：
1 class Solution { 2 public int findNthDigit(int n) { 3 int len = 1; 4 long count = 9; 5 int start = 1; 6 7 while (n > len * count) { 8 n -= len * count; 9 len += 1; 10 count *= 10; 11 start *= 10; 12 } 13 14 start += (n - 1) / len; 15 String s = Integer.toString(start); 16 return Character.getNumericValue(s.charAt((n - 1) % len)); 17 } 18 }
原文：http://www.cnblogs.com/mengchunchen/p/7883447.html