Question

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

Solution 1

class Solution:
  def findNthDigit(self, n: int) -> int:
    k, cnt = 1, 9
    while k * cnt < n:
      n -= k * cnt
      k += 1
      cnt *= 10
    num = 10 ** (k - 1) + (n - 1) // k
    idx = (n - 1) % k
    return int(str(num)[idx])

class Solution {
  public int findNthDigit(int n) {
    int k = 1, cnt = 9;
    while ((long) k * cnt < n) {
      n -= k * cnt;
      ++k;
      cnt *= 10;
    }
    int num = (int) Math.pow(10, k - 1) + (n - 1) / k;
    int idx = (n - 1) % k;
    return String.valueOf(num).charAt(idx) - '0';
  }
}

class Solution {
public:
  int findNthDigit(int n) {
    int k = 1, cnt = 9;
    while (1ll * k * cnt < n) {
      n -= k * cnt;
      ++k;
      cnt *= 10;
    }
    int num = pow(10, k - 1) + (n - 1) / k;
    int idx = (n - 1) % k;
    return to_string(num)[idx] - '0';
  }
};

func findNthDigit(n int) int {
  k, cnt := 1, 9
  for k*cnt < n {
    n -= k * cnt
    k++
    cnt *= 10
  }
  num := int(math.Pow10(k-1)) + (n-1)/k
  idx := (n - 1) % k
  return int(strconv.Itoa(num)[idx] - '0')
}

/**
 * @param {number} n
 * @return {number}
 */
var findNthDigit = function (n) {
  let k = 1,
    cnt = 9;
  while (k * cnt < n) {
    n -= k * cnt;
    ++k;
    cnt *= 10;
  }
  const num = Math.pow(10, k - 1) + (n - 1) / k;
  const idx = (n - 1) % k;
  return num.toString()[idx];
};

public class Solution {
  public int FindNthDigit(int n) {
    int k = 1, cnt = 9;
    while ((long) k * cnt < n) {
      n -= k * cnt;
      ++k;
      cnt *= 10;
    }
    int num = (int) Math.Pow(10, k - 1) + (n - 1) / k;
    int idx = (n - 1) % k;
    return num.ToString()[idx] - '0';
  }
}