# java 5ms solution beats 94.03%

• ``````  public  static int findNthDigit(int n) {
int digit = 1;//digit level, we start at one digit.
long counts = 9;//the number counts of current digit level,we start at one digit,there are 9 numbers(1-9)which is at one digit.
/**
* number [1-9] (there are 9 numbers)is of one digit,number[10-99](there are 90 numbers) is
* of two digits,number[100-999](there are 900 numbers) is of three digits,so first we should
* find what level(i mean which digits(one digit,two digit or so on)  by level) the nth digit locate,
* once we find the digit level, we achieve half the process,
*/
/**
*if n - digit * counts > 0,it means the nth digit is not at the current digit level,we should
* increase digit level to pass more number
*/

while (n - digit * counts > 0) {
//every time we pass the number at current digit level
n -= digit * counts;

digit++;
//counts are grow as follows,9,90,900,9000.....since the counts maybe overflow so i use long type
counts *= 10;
}
//after loop,the n means nth digits from the current baseNumber

//the base number is 1，10，100，1000，10000 and so on.
int baseNumber = (int)Math.pow(10, digit - 1);
//find the number where nth digit locate
int number  = (n -1) / digit + baseNumber;
//find the digit where nth digit locate at the number above
int mod = (n - 1 ) % digit;
return String.valueOf(number).charAt(mod) - '0';
}
``````

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