Nth Digit

题目地址：
https://leetcode.com/problems/nth-digit/description/

题目：

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.

解题思路：
link: https://discuss.leetcode.com/topic/59342/intuitive-solution-with-comments



代码：

/*    number of the digits at each "level"    1-9: 9 digits    10-99 : 90 * 2 = 180 digits    100-999 : 900 * 3 = 2700 digits    1000-9999 : 9000 * 4 = 36000 digits**/
public int findNthDigit(int n) {
    if(n < 1) return 0;
    if(n < 10) return n;
    int counter = 1;  //stores the level number    int base = 0;      //stores the biggest number from previous level    while(n > (9 * Math.pow(10,counter -1) * (counter))){
        base += 9 * Math.pow(10,counter -1);
        n -= (9 * Math.pow(10,counter -1) * (counter));
        counter++;
    }
    //target is the actual number that has nth digit    int target = base + ((n + counter - 1) / counter);  //to get the ceiling of n / counter    int offset = n % counter;
    offset = (offset == 0) ? 0:counter - offset;
    for(int i = 0; i < offset; i++){
        target = target / 10;
    }
    return target % 10;
}