LeetCode 400. Nth Digit

原题链接在这里：https://leetcode.com/problems/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.

题解：

先找到是在几位数的范围，再找具体是哪个数，再找是这个数的哪一位.

Note: count 是long, 如果是int的话, n = Integer.MAX_VALUE会overflow.

Time Complexity: O(m), m是最后len*count的位数. Space: O(1).

AC Java:

public class Solution {
    public int findNthDigit(int n) {
        int len = 1;
        long count = 9;
        int start = 1;
        
        //找到是几位数的范围
        while(n > len*count){
            n -= len*count;
            len += 1;
            count *= 10;
            start *= 10;
        }
        
        //找到具体的数
        start += (n-1)/len;
        
        //找到是这个数的第几位
        String s = Integer.toString(start);
        return s.charAt((n-1)%len) - '0';
    }
}