400 Nth Digit
Problem:
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.
Solutions:
public class Solution {
public int findNthDigit(int n) {
long base = 9;
int width = 1;
long count = 0; //number of digits already before the start the the range
int start = 1; //start number of the range
while (n > count + base * width) {
count += base * width;
base = base * 10;
width ++;
start = start * 10;
}
int diff = (int)(n - count - 1);
int number = start + diff / width;
int offset = width - diff % width - 1;
for (int i = 0; i < offset; i ++) {
number = number / 10;
}
return number % 10;
}
}