400. Nth Digit

最新推荐文章于 2022-12-05 20:19:33 发布

转载最新推荐文章于 2022-12-05 20:19:33 发布 · 86 阅读

本文介绍了一个算法问题，即如何找到无限整数序列1,2,3,4,5,6...中的第N位数字。通过递进的方式，文章详细解释了如何判断目标数字位于哪个长度级别的数字区间，并提供了具体的实现代码。

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.

class Solution {
public:
    int findNthDigit(int n) {
        if(n<=0)
            return -1;
        int index = n;
        int digits = 1;
        while(true)
        {
            long numbers =countOfIntegers(digits);
            if(index<numbers*digits)
                return digitAtIndex(index, digits);
            index -= digits*numbers;
            digits++;
        }
        return -1;
    }
private:
    long countOfIntegers(int digits)
    {
        if(digits == 1)
                return 10;
        int count = (int)std::pow(10, digits-1);
        return 9*count;
    }
    
    int digitAtIndex(int index, int digits)
    {
        int number = beginNumber(digits) + index/digits;
        int indexFromRight = digits - index%digits;
        for(int i=1; i<indexFromRight; ++i)
            number/=10;
        return number%10;
    }
    
    int beginNumber(int digits)
    {
        if(digits == 1)
            return 0;
        return (int)std::pow(10, digits-1);
    }
};