面试题 44. 数字序列中某一位的数字

输入：n = 3
输出：3

输入：n = 11
输出：0

class Solution:
    def findNthDigit(self, n: int) -> int:
        def get_bit_num():
            return 10 if p == 0 else 9 * pow(10, p) * (p + 1)

        if n < 10:
            return n
        p = count = 0
        while 1:
            count = get_bit_num()
            if n < count: break
            n -= count
            p += 1
        num = n // (p + 1) + pow(10, p)
        return int(str(num)[n % (p + 1)])

class Solution {
    public int findNthDigit(int n) {
        if (n < 10) return n;
        int pow = 0, count;
        while (true) {
            count = getBitNum(pow);
            if (n < count) break;
            n -= count;
            ++pow;
        }
        int num = n / (pow + 1) + (int) Math.pow(10, pow);
        return String.valueOf(num).charAt(n % (pow + 1)) - '0';
    }

    private int getBitNum(int pow) {
        if (pow == 0) {
            return 10;
        }
        return (int) (9 * Math.pow(10, pow) * (pow + 1));
    }
}

/**
 * @param {number} n
 * @return {number}
 */
var findNthDigit = function (n) {
  let i = 9;
  let a = 1;
  let remain = n;
  while (i * a < remain) {
    remain -= i * a;
    i *= 10;
    a++;
  }
  let b = remain % a;
  let res = 10 ** (a - 1) + ~~(remain / a);
  if (b === 0) {
    b = a;
    res--;
  }
  return res.toString()[b - 1];
};

class Solution {
public:
    int findNthDigit(int n) {
        int digit = 1;
        long long start = 0;
        long long count = 10;
        while (n > count) {
            n -= count;
            ++digit;
            start = start == 0 ? 10 : start * 10;
            count = 9 * start * digit;
        }
        long long num = start + n / digit;
        return to_string(num)[n % digit] - '0';
    }
};