400 Nth Digit

400. Nth Digit

1. Question

Find thenthdigit of the infinite integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...

Note: nis 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.

2. Implementation

(1) Math

class Solution {
    public int findNthDigit(int n) {
        // 记录target number的长度
        int digits = 1;
        // 这里要用Long防止overflow
        long count = 9;
        int startNum = 1;

        // Step 1: 找到第n个 digit所在的数的长度
        while (n > count * digits) {
            n -= count * digits;
            digits += 1;
            count *= 10;
            startNum *= 10;
        }

        // Step 2:找到target number
        int target = start +  (n - 1) / digits; //减一是因为start本身算一个数
        String s = Integer.toString(target); // 将target转成字符串方便下一步找到nth digit

        // Step 3: (n - 1) % digits是nth digit在target number的位置 
        return Character.getNumericValue(s.charAt((n - 1) % digits));
    }
}

3. Time & Space Complexity

时间复杂度O(logN), log的底是10， 空间复杂度O(1)