400. Nth Digit

#### 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.
``````

#### EXPLANATION:

1.首先需要算出是几位数。1位数是1-9，2位数是10-99.3位数就是100-999.在算这个的同时保存住上一个关键节点，比如1是9，2就是2*90+9，3就是3*900+2*90+9.

2.算出距离关键节点偏移了多少。

3.偏移的值除以几位数，就可以获取到具体的数了。

4.再算出第几位就可以了，取模就是具体的个位还是十位了。

#### SOLUTION:

``````public class Solution {
public int findNthDigit(int n) {
if (n <= 9) return n;
int sum = 0, index = 0, tmp = sum; double base = 0;
while (sum < n) {
index++;
tmp = sum;
base = Math.pow(10, index - 1);
sum += 9 * base * index;
}
int delta = (n - tmp - 1) / index;
int carry = (n - tmp - 1) % index;
int num = (int) (base + delta);
return (num + "").charAt(carry) - '0';
}
}
``````
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