README_EN.md

400. Nth Digit

中文文档

Description

Given an integer n, return the nth digit of the infinite integer sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...].

 

Example 1:

Input: n = 3
Output: 3

Example 2:

Input: n = 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.

 

Constraints:

• 1 <= n <= 231 - 1

Solutions

Python3

class Solution:
    def findNthDigit(self, n: int) -> int:
        bits, t = 1, 9
        while n > bits * t:
            n -= bits * t
            bits += 1
            t *= 10

        start = 10 ** (bits - 1) + (n // bits) - 1
        if n % bits == 0:
            return start % 10
        return int(str((start + 1))[(n % bits) - 1])

Java

class Solution {
    public int findNthDigit(int n) {
        int bits = 1, t = 9;
        while (n / bits > t) {
            n -= bits * t;
            ++bits;
            t *= 10;
        }
        int start = (int) Math.pow(10, bits - 1) + (n / bits) - 1;
        if (n % bits == 0) {
            return start % 10;
        }
        return String.valueOf(start + 1).charAt((n % bits) - 1) - '0';
    }
}

C++

class Solution {
public:
    int findNthDigit(int n) {
        int bits = 1, t = 9;
        while (n / bits > t)
        {
            n -= bits * t;
            ++bits;
            t *= 10;
        }
        int start = pow(10, bits - 1) + (n / bits) - 1;
        if (n % bits == 0) return start % 10;
        return to_string(start + 1)[(n % bits) - 1] - '0';
    }
};

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