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.这道题的解题思路分下面三步:
1、找到nth number 所在的数的长度
2、找到nth number 在这个长度所有数的哪一个
3、找到nth number 是这个数的第几位
代码如下:
public class Solution {
public int findNthDigit(int n) {
int len = 1;
long count = 9;
int start = 1;
while (n > len * count) {
n -= len * count;
len ++;
count *= 10;
start *= 10;
}
start += (n-1) / len;
String numstr = Integer.toString(start);
System.out.println(numstr);
System.out.println(n);
return numstr.charAt((n - 1) % len) - '0';
}
}
这里注意count 需要是long,因为在长度为8时,所含的数有90000000个,超过了Integer.MAX_VALUE。