# Annotated C++ solution

• ``````class Solution {
public:
int findNthDigit(int n) {
int digits[] = {0,1,2,3,4,5,6,7,8,9};
// trivial case first
if( n < 10 ){
return digits[n];
}
// our sequence is just a series of non-delimited numbers
// position _n_ in this sequence points to a certain digit of some number
// these are 9 one-digit numbers in the trivial case (call it _order_=0)
//  then they turn to 90 two-digit numbers ranging in [10-99] (call it _order_=1)
//  after that they turn to 900 three-digit numbers from [100-999] range (_order_=2)
// let us first identify which _order_ we deal with for the given _n_ and
//  subtract from _n_ the lengths of each of the previous orders
int order = 1;
for( ; ; order++){
unsigned long long shift = (pow(10,order) - pow(10,order-1))*order;
if( n <= shift ) break;
n -= shift;
}
// now _n_ is pointing to a digit in a sequence of same width numbers
// let us identify the number (keep in mind that _n_ start counting from 1)
int num = (n-1) / order + pow(10,order-1) ;
//  and the position in this number
int pos = order-1 - (n-1) % order;
// so we identified the number we are on and digit's position in this number
while( pos-- ) num /= 10;
// return it
return digits[num%10];
}
};``````

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