Question

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

就是统计位数,然后注意越界的问题,所以应该用long.

Code

class Solution {
public:
int findNthDigit(int n) {
if (n < 10)
return n;
int i = 1;
long pre = 0;
while (1) {
long value = i * 9 * pow(10, i - 1);
if (n > value) {
pre += value;
i++;
}
else {
break;
}
}
long remain = n - pre;
long y = remain / i;
long z = remain % i;
i--;
long start;
if (z > 0)
start = pow(10, i) + y;
else
start = pow(10, i) + y - 1;
stringstream ss;
ss << start;
string str;
ss >> str;
if (z > 0)
return (str[z - 1]) - 48;
else
return (str[str.length() - 1]) - 48;
}
};

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