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.
Solution：
思路：
Straight forward way to solve the problem in 3 steps:
find the length of the number where the nth digit is from
find the actual number where the nth digit is from
find the nth digit and return
Time Complexity: O(N) Space Complexity: O(1)
Solution Code：
class Solution {
public int findNthDigit(int n) {
int len = 1;
long count = 9;
int start = 1;
while (n > len * count) {
n -= len * count;
len += 1;
count *= 10;
start *= 10;
}
start += (n - 1) / len;
String s = Integer.toString(start);
return Character.getNumericValue(s.charAt((n - 1) % len));
}
}
