400. Nth Digit

题目

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.

我的解法

public class Solution {    public int findNthDigit(int n) {        // 数字的位数        int len = 1;        double sum = 0;        // 当加上len长度所有数字后sum>=n时，说明在此len长度内        while(sum < n){            sum += len*9*Math.pow(10, len-1);            len++;        }        len = len - 1;        int dis = (int)(n - (sum - len*9*Math.pow(10, len-1)));        int num = (int)(Math.pow(10, len - 1) + Math.floor((dis-1)/len));        return (int)Integer.toString(num).charAt((dis-1)%len) - '0';    }}

算法分析：累加每个len长度内数字的个数，来发现n非区间；然后继续缩小范围，来求解。