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.
思路：
这道题其实停费脑子的，先要找到区间，然后找到位置
class Solution {
public int findNthDigit(int n) {
if(n<10) return n;
long num = 9;
int count = 1;
while(n > count * num){
n -= num * count;
count ++;
num *= 10;
}
long res = num/9 + (n-1)/count;
return Integer.parseInt(String.valueOf(res).substring((n-1)%count, (n-1)%count+1));
}
}
本站原创文章皆遵循“署名—非商业性使用—相同方式共享 4.0 协议 (CC BY-NC-SA 4.0)”。共享、演绎请保留以下标注：
原文作者：Jake Tao，来源：「400. Nth Digit」
